Answer
$y^{''}=\large\frac{\cos^2 y\;\cos x+\sin^2 x\;\sin y}{\cos^3 y}$
Work Step by Step
$\sin y+\cos x=1$ ____(1)
Differentiating (1) implicitly with respect to $x$
$\cos y\; \cdot y^{'}-\sin x=0$
Solving for $y^{'}$
$y^{'}=\large\frac{\sin x}{\cos y}$ ____(2)
Differentiating (2) implicitly with respect to $x$
$y^{''}=\large\frac{d}{dx}\left(\frac{\sin x}{\cos y}\right)$
$y^{''}=\large\frac{\cos y\;(\sin x)^{'}-\sin x\;(\cos y)^{'}}{\cos^2 y}$
$y^{''}=\large\frac{\cos y\;\cos x+\sin x\;\sin y\; \frac{dy}{dx}}{\cos^2 y}$
From (2)
$y^{''}=\large\frac{\cos y\;\cos x+\sin x\;\sin y\; \left(\large\frac{\sin x}{\cos y}\right)}{\cos^2 y}$
$y^{''}=\large\frac{\cos^2 y\;\cos x+\sin^2 x\;\sin y}{\cos^3 y}$
Hence $y^{''}=\large\frac{\cos^2 y\;\cos x+\sin^2 x\;\sin y}{\cos^3 y}$.