Given w (x, y, z) = x² +y² +y², 
x = e^t, y = e^t sin t, z = e^t cos t
\(\frac { \partial u }{ \partial x } \) = 2x; \(\frac { \partial u }{ \partial y} \) = 2y; \(\frac { \partial u }{ \partial z} \) = 2z
\(\frac { \partial u }{ \partial x } \) = 2e^t
\(\frac { \partial u }{ \partial y} \) = 2e^t sin t
\(\frac { \partial u }{ \partial z} \) = 2e^t cos t
\(\frac{dx}{dt}\) = e^t
\(\frac{dy}{dt}\) = e^t cas t + sin t e^t
⇒ \(\frac{dz}{dt}\) = et (- sin t ) + cos t et
By chain rule
\(\frac { dw }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { dw }{ \partial y } .\frac { dy }{ dt } +\frac { \partial w }{ \partial z } .\frac { dz }{ dt } \)
∴ \(\frac { dw }{ dt } \) = 2e^t(e^t) + 2e^t sin t (e^t cos t +sin t e^t) - e^t sin t + 2e^t cas t (e^t cos t - e^t sin t )

= e^2t [2 + 2] = 4e^2t

Let F(x, y) = x² – 2y² + 2xy and x(t) = cost, y(t) = sint
Then F(x, y) = cos² t - 2sin² t + 2cos t sin t  and thus F has becomes a function of one variable t. So by using chain rule, we see that
\(\frac { dF }{ dt } \) = 2 cos t(-sin t) -4 sin t cos t 2 (-sin² t + cos² t).
= -6 cos t sin t +2 ( -sin² t + cos² t)
On the other hand if we calculate
\(\frac { \partial F }{ \partial x } \frac { \partial x }{ \partial t } +\frac { \partial F }{ \partial y } \frac { \partial y }{ \partial t } \) = (2x+2y)\(\frac { d x }{ dt } \)+(2x - 4y) \(\frac { dy }{ dt } \)
= 2(cos t + sin t)(-sin t) + 2(cos t - 2sin t)(cos t)
= -6 cos t sin t +2 ( -sin² t + cos² t)
= \(\frac { dF }{ dt } \)

g(x, y) = log (5x + 3y)
\({ g }_{ x }=\frac { 1 }{ 5x+3y } (5)=\frac { 5 }{ 5x+3y } \)
\({ g }_{ y }=\frac { 1 }{ 5x+3y } (3)=\frac { 3 }{ 5x+3y } \)
\({ g }_{ xy }=\frac { \partial }{ \partial x } ({ g }_{ y })\)
= 3(-1)(5x + 3y)^-2 (5)
\(=\frac { -15 }{ { (5x+3y) }^{ 2 } } \)
\({ g }_{ xx }=\frac { \partial }{ \partial x } ({ g }_{ x })=\frac { -5 }{ ({ 5x+3y) }^{ 2 } } (5)\)
\(=\frac { -25 }{ ({ 5x+3y) }^{ 2 } } \)
\({ g }_{ yy }=\frac { \partial }{ \partial y } ({ g }_{ y })=\frac { -3 }{ ({ 5x+3y) }^{ 2 } } (3)\)
\(=\frac { -9 }{ ({ 5x+3y) }^{ 2 } } \)
\({ g }_{ yx }=\frac { \partial }{ \partial y } ({ g }_{ x })=\frac { -5 }{ 5x+3y } (3)\)
\(=\frac { -15 }{ ({ 5x+3y) }^{ 2 } } \)

g(x, y) = xe^y + 3x²y
g_x = e^y + 6xy
gy = xey + 3x²
\({ g }_{ xy }=\frac { \partial }{ \partial x } ({ g }_{ y })={ e }^{ y }+6x\)
\({ g }_{ xx }=\frac { \partial }{ \partial x } ({ g }_{ xy })=0+6y=6y\)
\({ g }_{ yy }=\frac { \partial }{ \partial y } ({ g }_{ y })={ xe }^{ y }\)
\({ g }_{ xy }=\frac { \partial }{ \partial y } ({ g }_{ x })\) = e^y + 6

Given f(x, y) = cos (x² - 3xy)
f_x = - sin (x² - 3xy) [2x - 3y]
= (3y - 2x) sin (x² - 3xy)
f_y = - sin (x² - 3xy) (-3x)
= 3x sin (x² - 3xy)
∴ f_xy = \(\frac { \partial }{ \partial x } ({ f }_{ y })\)
= 3 [x cos (x² - 3xy) (2x - 3y) + sin (x² - 3xy) (1)]
= 3[(2x²- 3xy)cos(x² - 3xy) + sin (x² - 3xy)] ... (1)
∴ f_yx  = \(\frac { \partial }{ \partial y} ({ f }_{ x })\)
= (3y- 2x) cos(x² - 3xy) (- 3x) + sin (x² - 3xy) (3)
= 3[(2x²- 3xy)cos(x² - 3xy) + sin (x² - 3xy)] ... (2)
From (1) and (2),f_xy =1.

Note that the function f takes value 1 on the x, y-axes and 0 everywhere else on R². So let us calculate
\(\frac { \partial f }{ \partial x } (0,0)\) = \(\underset { h\longrightarrow 0 }{ lim } \) \(\frac { f\left( 0+h,0 \right) -f(0,0) }{ h } =\underset { h\longrightarrow 0 }{ lim } \frac { 1-1 }{ h } =0;\)
\(\frac { \partial f }{ \partial y } (0,0)\) = \(\underset { k\longrightarrow 0 }{ lim } \) \(\frac { f\left( 0,0+k \right) -f(0,0) }{ k } =\underset { k\longrightarrow 0 }{ lim } \frac { 1-1 }{ k } =0\)
This completes (i).

\(\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}cos=\left( \frac { { e }^{ x }siny }{ y } \right) \) = \(cos\left( { e }^{ 0 }\frac { siny }{ y } \right) \)
= cos[(1)(1)] = cos (1) \(\left[ \because \begin{matrix} lim \\ y\rightarrow 0 \end{matrix}\frac { siny }{ y } =1 \right] \)

Given u (x, y) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ \sqrt { x+y } } \)
\(u({ \lambda }x,{ \lambda }y)=\frac { { \lambda }^{ 2 }{ x }^{ 2 }+{ { \lambda } }^{ 2 }{ y }^{ 2 } }{ \sqrt { { \lambda }x+{ \lambda }y } } \)
= \(\frac { { { \lambda } }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) }{ \sqrt { { \lambda } } (\sqrt { x+y } ) } \)
= \({ { \lambda } }^{ 2-\frac { 1 }{ 2 } }u(x,y)\)
= \({ { \lambda } }^{ \frac { 3 }{ 2 } }u(x,y)\)
∴ u (x, y) is a homogeneous function of degree \(\frac32\)
∴ By Euler's theorem,
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } \) = n.u ≍ \(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 3 }{ 2 } u\)
Hence, proved.

Given  W(x, y, z) = x² - xy + 3 sin z, x, y, z ∈ R
W(x₀, y₀, z₀) = W (2, -1, 0)
= 2²-2(-1)+3 sin 0
\(\frac { \partial W }{ \partial x } \) = 2x - y + 0 = 2x - y
\(\left( \frac { \partial W }{ \partial x } \right) _{(2, -1, 0)}\)= 2(2)-(-1) = 5
\(\frac { \partial W }{ \partial y } \) = 0-x + 0 = -x
\(\left( \frac { \partial W }{ \partial y } \right) _{(2, -1, 0)}\) = -2
\(\frac { \partial W }{ \partial z } \) = 0 - 0 + 3 cos z
 \(\left( \frac { \partial W }{ \partial z} \right) _{(2, -1, 0)}\)= 3 cos 0 = 3(1) = 3.
Linear approximation is given by
L(x, y, z) = w(x₀, y₀, z₀) + \(\left( \frac { \partial W }{ \partial x } \right) _{(x_0, y_0, z_0)}\) (x - x₀) + \(\left( \frac { \partial W }{ \partial y} \right) _{(x_0, y_0, z_0)}\) (y - y₀) + \(\left( \frac { \partial W }{ \partial z} \right) _{(x_0, y_0, z_0)}\)(z₀ - z₀)
∴ L(x,y,z) = 6 + 5(x - 2) - 2(y+1) + 3(z - 0)
= 6 + 5x - 10 - 2y - 2 + 3z
L (x, y, z) = 5x - 2y + 3z - 6

Givenz (x, y) = x²y+ 3xy⁴, x, y ∈ R.
z (x_o, y_o) = z (2, -1) = 2²(-1) + 3(2)(-1)4
= -4 + 6 = 2
\(\frac { \partial z }{ \partial x } \) = 2xy + 3y⁴
\(\left( \frac { \partial z }{ \partial x } \right) \)_{(2, -1)} = 2 (2) (-1) + 3 (1)4
= 4+3 = -1
\(\frac { \partial v }{ \partial y} \) = x² + 3x (4y³)
= x2 + 12xy3
\(\left( \frac { \partial z }{ \partial y } \right) \)_{(2, -1)} = 22+ 12 (2)(-1)³
= 4 -24 = -20
Linear approximate is given by
L(x, y) = z(x_o, y_o) + \(\left( \frac { \partial z }{ \partial x} \right) \)(x_o, y_o) (x - x₀) 
L(x, y) = z(x_o, y_o) + \(\left( \frac { \partial z }{ \partial z} \right) \)(x_o, y_o) (x - x₀)
∴ L(x, y) = 2-1 (x-2) - 20(y + 1)
= 2 - x + 2 - 20y - 20
= -x - 20y - 16
= - (x + 20y + 16)