Explanation

(a)
Given equation is
$y - e^{xy} + x = 0$
$\Rightarrow \ \ \ e^{xy} = x + y$
On taking log on both sides, we get
$\log(x + y) = xy$
On differentiating w.r.t. $x$, we get
$\frac{1}{x + y} + \left( 1 + \frac{dy}{dx} \right) = x\frac{dy}{dx} + y$
$\Rightarrow \ \frac{dy}{dx}\left( \frac{1}{y + x} - x \right) = y - \frac{1}{y + x}$
$\Rightarrow \ \ \frac{dy}{dx} = \frac{y(y + x) - 1}{1 - x(y + x)}$
Since, the curve has a vertical tangent
$\therefore\ \ \frac{dy}{dx} = \infty\ \ \Rightarrow \ \ 1 - x(x + y) = 0$
Which is satisfied by the point (1, 0)