Explanation

(b)
Given, $z = y + f(v)$ …(i)
Where $v = \left( \frac{x}{y} \right)$
On differentiating partially Eq. (i) w.r.t. $x$, we get
$\ \ \frac{\partial z}{\partial x} = f^{'}\left( \frac{x}{y} \right).\left( \frac{1}{y} \right)$
$\Rightarrow v\frac{\partial z}{\partial x} = f^{'}\left( \frac{x}{y} \right).\left( \frac{1}{y} \right)\left( \frac{x}{y} \right) = f^{'}\left( \frac{x}{y} \right)(\frac{x}{y^{2}})\ \ \ \ \$…(ii)
Now, differentiating partially Eq. (i) w.r.t. $y$,$\$we get
$\frac{\ \partial z}{\partial y} = 1 + f^{'}\left( \frac{x}{y} \right)\left( \frac{- x}{y^{2}} \right)$ …(iii)
On adding Eqs. (ii) and (iii), we get
$v\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = f^{'}\left( \frac{x}{y} \right)\left( \frac{x}{y^{2}} \right) + 1 + f^{'}\left( \frac{x}{y} \right)\left( \frac{- x}{y^{2}} \right) = 1$