Explanation

(a)
Given, $x^{y} = e^{x - y}$
Taking log on both sides, we get
$y\log{x = x - y}$
On differentiating w.r.t. $x$, we get
$\ \ \frac{y}{x} + log\ x\frac{dy}{dx} = 1 - \frac{dy}{dx}$
$\Rightarrow \ \frac{dy}{dx}(1 + \log{x) = 1 - \frac{y}{x}}$
$\Rightarrow \left( \frac{dy}{dx} \right)_{(1,1)}(1 + \log{1) = 1 - \frac{1}{1}}$
$\Rightarrow {\ \ \ \left( \frac{dy}{dx} \right)}_{(1,1)} = 0$