Explanation

(a)
Given, $y = \cos{x + x\cos{y = \pi}}$
On differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}\cos{x + y( - \sin{x) + x( - \sin{y)\frac{dy}{dx} + \cos{y = 0}}}}$
$\Rightarrow \frac{dy}{dx} = \frac{y\sin{x - \cos y}}{\cos{x - x\sin y}}$
Again differentiating both sides w.r.t. $x$, we get
$\frac{d^{2}y}{dx^{2}} =$
$\frac{\left\lbrack \begin{array}{r}
\left( \cos x - x\sin y \right)\left( y\cos{x + \sin{x\frac{dy}{dx} + \sin{y\frac{dy}{dx}}}} \right) \\
- (y\sin{x - \cos{y)\left( - \sin{x - \sin{y - x\cos{y\frac{dy}{dx}}}} \right)}}
\end{array} \right\rbrack}{(\cos{x - x\sin{y\left. \ \ \right)^{2}}}}$
At $x = 0$,
$f^{''}(0) = \frac{1(y + \sin{y) - ( - 1)( - \sin y)}}{(1 - 0)^{2}} = y$
As $y\cos{x + x\cos{y = \pi}}$
$\therefore At\ x = 0 \Rightarrow y = \pi$
Hence, $f^{''}(0) = \pi$