Answer
$y"=\frac{cosxcos^2y+sin^2xsiny}{cos^3y}$
Work Step by Step
Find y' by taking the derivative of both sides of the equation:
$cosy\times y'-sinx=0$
$y'=\frac{sinx}{cosy}$
Find y":
$y"=\frac{(cosy)(cosx)-(sinx)(-siny\times y')}{cos^2y}$
Plug y' into y":
$y"=\frac{cosxcosy+sinxsiny\times\frac{sinx}{cosy}}{cos^2y}$
Simplify:
$y"=\frac{cosxcos^2y+sin^2xsiny}{cos^3y}$
