Given (x,y) = x³ - 2x²y + 3xy² + y³ ...(1)
f(tx, ty) = (tx)³ - 2(tx)² (ty) + 3 (tx) (ty)² + (ty)³
= t³ x³ - 2t² x²ty + 3txt²y²+ t³y³
= t³ (x³ - 2x²y + 3xy² +y³)
f(tx, ty) = t³.f(x, y)
∴ f is a homogeneous function and its degree is 3.
Differentiate (1) partially with respect to 'x' and 'y' we get
\(\frac { \partial f }{ \partial x } { =3 }^{ 2 }-4xy+3{ y }^{ 2 }\)
\(\Rightarrow x\frac { \partial f }{ \partial x } ={ 3x }^{ 2 }-4xy+{ 3xy }^{ 2 }\)  ...(2)
\(\frac { \partial f }{ \partial y } =-{ 2x }^{ 2 }+6xy+3{ xy }^{ 2 }\)
\(\Rightarrow y\frac { \partial f }{ \partial y } =-2{ x }^{ 2 }+{ 6xy }^{ 2 }+{ 3y }^{ 3 }\) ...(3)
Adding (2) and (3) we get,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 3x² - 4x²y + 3x²y- 2x²y + 6xy² + 3y³
= 3x² - 6x²y + 9xy² + 3y³
= 3 (x³ - 2x²y - 3xy² +y³)
= 3f [using (1)]
∴ \(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 3f = nf where 3 is the degree of (x, y)
Hence Euler's theorem is verified.

W(x, y, z) = xy + yz + zx, x =u -v, y = uv, z = u + v; y = uv; z = u
\(\frac { \partial W }{ \partial x } \) = y + z; \(\frac { \partial W }{ \partial y } \) = x + z
∴ \(\frac { \partial W }{ \partial x } \) = uv + u +  v;
\(\frac { \partial W }{ \partial y} \) = u - v + u + v;
\(\frac { \partial W }{ \partial z} \) = uv +  u - v
\(\frac { dx }{ du } =1;\frac { dy }{ du } =v;\frac { dz }{ du } =1\)
\(\frac { dx }{ dv } =1;\frac { dy }{ dv } =v;\frac { dz }{ dv} =1\)
By chain rule
\(\frac { \partial W }{ \partial u } =\frac { \partial w }{ \partial x } .\frac { dx }{ du } +\frac { \partial w }{ \partial y } .\frac { dy }{ du } +\frac { \partial w }{ \partial z } .\frac { dz }{ du } \)
= (uv +u +v) (1) +2u (v) + (uv +u - v)(1)

\(\frac { \partial W }{ \partial u } \) = 4uv + 2u = 12u (2v + 1)
\({ \left( \frac { \partial W }{ \partial u } \right) }_{ \left( \frac { 1 }{ 2 } ,1 \right) }\) = 2 x \(\frac12\) (2+ 1) = 1(2+ 1) = 3
= (uv + u + v) (-1) + (2u) (u) + (uv + u - v)(1)

= 2u² - 2v = 2 (u² - v)
∴  \({ \left( \frac { \partial W }{ \partial v } \right) }_{ \left( \frac { 1 }{ 2 } ,1 \right) }\) = \(2\left( \frac { 1 }{ 4 } -1 \right) =2\left( -\frac { 3 }{ 4 } \right) =-\frac { 3 }{ 2 } \)

Given U (x, y) = e^x sin y ; x = st² ; y = s²t
\(\frac { \partial U }{ \partial x } \) = e^x sin y ; \(\frac { \partial U }{ \partial y } \) = e^x cos y
\(\frac { \partial U }{ \partial x } \) = \({ e }^{ { st }^{ 2 } }\) sin (s²t)
\(\frac { \partial U }{ \partial y } \) = \({ e }^{ { st }^{ 2 } }\) cos (s²t)
\(\frac{dx}{dt}\) = 2st; \(\frac{dy}{dt}\) = s²
\(\frac{dx}{ds}\) = t²; \(\frac{dy}{ds}\) = 2 st
By chain rule
\(\frac { dU }{ ds } =\frac { \partial U }{ \partial x } .\frac { dx }{ ds } +\frac { \partial U }{ \partial y } .\frac { dy }{ ds } \)
= \({ e }^{ { st }^{ 2 } }\). sin (s²t) (t²) + \({ e }^{ { st }^{ 2 } }\) cos(s^2t).(2^st)
∴ \({ \left( \frac { \partial U }{ \partial s } \right) }_{ (s=t=1) }\) = e¹ sin (1) + 2e¹ cos (1)
= e [sin (1) + 2 cos (1)] and
\(\frac { dU }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { \partial u }{ \partial y } .\frac { dy }{ dt } \)
=  \({ e }^{ { st }^{ 2 } }\) . sin (s²t)(2^st) + \({ e }^{ { st }^{ 2 } }\) cos (s²t). (s²)
∴ \({ \left( \frac { \partial U }{ \partial t } \right) }_{ (s=t=1) }\) = 2e¹ sin (1) + e¹ cos (1)
= e [2 sin (1) + cos (1)]

\(\frac { \partial z }{ \partial x } =x.\frac { 1 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } (y)+{ tan }^{ -1 }(xy)\)
\(\frac { \partial z }{ \partial y } =x.\frac { 1 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)(x)
\(\frac { \partial z }{ \partial z } =\frac { xy }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \) + tan^-1 (xy)
\(\frac { \partial z }{ \partial y } =\frac { x^2 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)
\(\frac { \partial z }{ \partial y } =\frac { x^2 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)
\(\frac { \partial z }{ \partial x } =\frac { { t }^{ 2 }{ se }^{ t } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } +{ tan }^{ -1 }({ t }^{ 2 }{ se }^{ 2 })\)
\(\frac { \partial z }{ \partial y } =\frac { { t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } \)
\(\frac { dx }{ dt } =2t;\frac { dy }{ dt } ={ s.e }^{ t }\)
Also, \(\frac { dx }{ ds } =0;\frac { dy }{ ds } ={ e }^{ t }\)
By chain rule
\(\frac { dw }{ ds } =\frac { \partial z }{ \partial x } .\frac { dx }{ ds } +\frac { \partial z }{ \partial y } .\frac { dy }{ ds } \)
= \(\frac { { t }^{ 2 }{ se }^{ t } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } (0)+\frac { { t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } { (e }^{ t })=\frac { { e }^{ t }{ t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } \)
\(\therefore { \left( \frac { \partial z }{ \partial s } \right) }_{ s=t=1 }=\frac { e(1) }{ 1+{ e }^{ 2 } } =\frac { e }{ 1+{ e }^{ 2 } } \)
By chain rule
= \(\frac { dz }{ dt } =\frac { \partial z }{ \partial x } .\frac { dx }{ dt } +\frac { \partial z }{ \partial y } .\frac { dy }{ dt } \)
\(\therefore { \left( \frac { \partial z }{ \partial t } \right) }_{ (s=t=1) }=\frac { e }{ 1+{ e }^{ 2 } } =\frac { e }{ 1+{ e }^{ 2 } } \)
= \(\frac { 2e+e }{ 1+{ e }^{ 2 } } =\frac { 3e }{ 1+{ e }^{ 2 } } \)+ 2 tan^-1(e)

Given w (x, y) = xy + sin (xy)
\(\frac { \partial w }{ \partial x } \) = y (1) + (cos (xy) [y (1)]
= y + y cos (xy)
\(\frac { \partial w }{ \partial y } \) = x (1) + cos (xy) (x)
= x + x cos (xy)
\(\frac { { \partial }^{ 2 }w }{ \partial y\partial x } \) = \(\frac { { \partial } }{ \partial { y } } \left( \frac { \partial w }{ \partial x } \right) \)
= 1 + y (- sin (xy)) (x) + cos (xy)
= 1 -xy sin (xy) + cos (xy) ... (1)
\(\frac { { \partial } }{ \partial { x } } \left( \frac { \partial w }{ \partial y } \right) \)
= 1 + x (- sin (xy)) (y) + cos (xy)
= 1 - xy sin (xy) + cos (xy) ... (2)
∴ From (1) and (2),
\(\frac { { \partial }^{ 2 }w }{ \partial x\partial y } =\frac { { \partial }^{ 2 }w }{ \partial x\partial y } \)