Let \(y=\sqrt { \cfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3x }^{ 2 }+4x+5 } } \) 
= \(\left| \cfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3x }^{ 2 }+4x+5 } \right| ^{ \frac { 1 }{ 2 } }\) 
Taking logarithm on both sides,
\(logy=\cfrac { 1 }{ 2 } \left[ log\left( x-3 \right) +log\left( { x }^{ 2 }+4 \right) -log\left( { 3x }^{ 2 }+4x+5 \right) \right] \)\({[\because \log a b} =\log a+\log b \text { and } \log \frac{a}{b} =\log a-\log b]\) 
Differentiating with respect to x
\(\cfrac { 1 }{ y } .\cfrac { dy }{ dx } =\cfrac { 1 }{ 2 } \left[ \cfrac { 1 }{ x-3 } +\cfrac { 2x }{ { x }^{ 2 }+4 } -\cfrac { 6x+4 }{ { 3x }^{ 2 }+4x+5 } \right] \)
\(\cfrac { dy }{ dx } =\cfrac { 1 }{ 2 } \sqrt { \cfrac { \left( x-3 \right) \left( { x }^{ 2 }+4 \right) }{ { 3x }^{ 2 }+4x+5 } } \)\( \left[ \cfrac { 1 }{ x-3 } +\cfrac { 2x }{ { x }^{ 2 }+4 } -\cfrac { 6x+4 }{ { 3x }^{ 2 }+4x+5 } \right] \)

\(f(x)=|x|=\left\{\begin{array}{cll} x & \text { if } & x \geq 0 \\ -x & \text { if } & x<0 \end{array}\right.\)
\(\mathrm{L}\left[f^{\prime}(0)\right]=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\)
\(=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}, x=0-h\)
\(=\lim _{h \rightarrow 0} \frac{f(-h)-f(0)}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{|-h|-|0|}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{|-h|}{-h}\)
\(=\lim _{h \rightarrow 0} \frac{h}{-h}\)
\(=\lim _{h \rightarrow 0}(-1)=-1\)
\(\mathrm{R}\left[f^{\prime}(0)\right]=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)
\(=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0}, x=0+h\)
\(=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{|h|-|0|}{h}\)
\(=\lim _{h \rightarrow 0} \frac{|h|}{h}\)
\(=\lim _{h \rightarrow 0} \frac{h}{h}\)
\(=\lim _{h \rightarrow 0} 1=1\)
\(\text { Here, } \mathrm{L}\left[f^{\prime}(0)\right] \neq \mathrm{R}\left[f^{\prime}(0)\right]\)
\(\therefore\) f(x) is not differentiable at x = 0

\(L\left| f(x) \right| _{ x=1 }=\underset { x\rightarrow { 1 }^{ - } }{ lim } f(x)\\ =\underset { h\rightarrow 0 }{ lim } f\left( 1-h \right) ,x=1-h\\ =\underset { h\rightarrow 0 }{ lim } \left[ 5\left( 1-h \right) -4 \right] \\ =5(1)-4=1\) 
\(R\left[ f\left( x \right) \right] _{ x=-1 }=\underset { x\rightarrow { 1 }^{ + } }{ lim } f(x)\\ =\underset { h\rightarrow 0 }{ lim } f\left( 1+h \right) ,x=1+h\\ =\underset { h\rightarrow 0 }{ lim } \left[ 4(1+h)^{ 2 }-3\left( 1+h \right) \right] \\ =4\left( 1 \right) ^{ 3 }-3(1)\\ =4-3=1\)
Now.\(f(1)=5(1)-4=5-4=1\) 
\(\therefore \underset { x\rightarrow { 1 }^{ - } }{ lim } f(x)=\underset { x\rightarrow { 1 }^{ + } }{ lim } f(x)=f\left( 1 \right) \) 
f(x) is  continuous at x = 1

Let y = \(\sqrt { \frac { (x-1)(x-2) }{ (x-3)({ x }^{ 2 }+x+1) } }\)
log y = \(\frac { 1 }{ 2 } \)[log (x - 1) + log (x - 2) -log (x - 3) -log (x² + x + 1]
\( \frac{1}{y} \frac{d y}{d x}=\frac{1}{2}\left[\begin{array}{l} \left.\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}\right] \\ -\frac{1}{x^2+x+1}(2 x+1) \end{array}\right]\)
⇒ \(\frac { dy }{ dx } =\frac { 1 }{ 2 } \sqrt { \frac { (x-1)(x-2) }{ (x-3)(x^{ 2 }+x+1) } } \left[ \frac { 1 }{ x-1 } +\frac { 1 }{ x-2 } -\frac { 1 }{ x-3 } -\frac { (2x+1) }{ { x }^{ 2 }+x+1 } \right] \)

Given \(\sqrt { 1+y } +y\sqrt { 1+x } =0\)
\(\Rightarrow \ x\sqrt { 1+y } = -y\sqrt { 1+x } \)
Squaring on both sides
\(\Rightarrow { x }^{ 2 }(1+y)={ y }^{ 2 }(1+x)\)
\(\Rightarrow { x }^{ 2 }+{ x }^{ 2 }y={ y }^{ 2 }+x{ y }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }-{ y }^{ 2 }=x{ y }^{ 2 }-{ x }^{ 2 }y\)
\(x y(x-y)=(y-x)(y+x)\)
\(x y=\frac{-(x-y)(y+x)}{x-y}=-y-x\)
\(x y+y=-x\)
\(y(x+1)=-x \Rightarrow y=\frac{-x}{x+1}\)