Answer
\[y^{'}=\frac{y\sec^2\left(\frac{x}{y}\right)-y^2}{y^2+x\sec^2\left(\frac{x}{y}\right)}\]
Work Step by Step
$\tan\left(\frac{x}{y}\right)=x+y$ ___(1)
Differentiating (1) implicitly with respect to $x$
$\sec^2\left(\frac{x}{y}\right)\large\left[\frac{(x)^{'}y-x(y)^{'}}{y^2}\right]$=$1+y^{'}$
$\sec^2\left(\frac{x}{y}\right)\large\left[\frac{y-xy^{'}}{y^2}\right]$=$1+y^{'}$
$\frac{1}{y}\sec^2\left(\frac{x}{y}\right)-\frac{x}{y^2}\sec^2\left(\frac{x}{y}\right)y^{'}=1+y^{'}$
$\frac{1}{y}\sec^2\left(\frac{x}{y}\right)-1=y^{'}\left[1+\frac{x}{y^2}\sec^2\left(\frac{x}{y}\right)\right]$
$y^{'}=\Large\frac{\frac{\sec^2\left(\frac{x}{y}\right)-y}{y}}{\frac{y^2+x\sec^2\left(\frac{x}{y}\right)}{y^2}}$
$y^{'}=\Large\frac{y\sec^2\left(\frac{x}{y}\right)-y^2}{y^2+x\sec^2\left(\frac{x}{y}\right)}$
Hence $y^{'}=\Large\frac{y\sec^2\left(\frac{x}{y}\right)-y^2}{y^2+x\sec^2\left(\frac{x}{y}\right)}$.
