Given that, $\left(\frac{d y}{d x}\right)=\frac{1}{\left(\frac{d x}{d y}\right)}$ or $\frac{d y}{d x}=\left(\frac{d x}{d y}\right)^{-1} \ldots$(i)
Differentiating Eq. (i) on both sides w.r.t. $x$, we get
$\frac{d^2 y}{d x^2}=-\left(\frac{d x}{d y}\right)^{-2}\left\{\frac{d}{d x}\left(\frac{d x}{d y}\right)\right\}$
$\Rightarrow \quad \frac{d^2 y}{d x^2}=-\left(\frac{d x}{d y}\right)^{-2}\left\{\frac{d}{d x}\left(\frac{d x}{d y}\right) \frac{d y}{d x}\right\}$ [from chain rule]
$\Rightarrow \quad \frac{d^2 y}{d x^2}=\left(\frac{d x}{d y}\right)^{-2}\left\{\frac{d^2 x}{d y^2} \cdot \frac{d x}{d y}\right\}$
$\Rightarrow \quad \frac{d^2 y}{d x^2}=-\left(\frac{d y}{d x}\right)^2 \frac{d^2 x}{d y^2} \Rightarrow \frac{d^2 x}{d y^2}+\left(\frac{d y}{d x}\right)^3 \frac{d^2 x}{d y^2}=0$
Comparing with given expression, we get
$k=0$
Now, $e^{k f(x)}-k f(x)=e^{0 \times f(x)}-0 \times f(x)$
$e^{k f(x)}-k f(x)=e^0-0=1 \quad\left[\because e^0=1\right]$