Find an expression for $\frac{dy}{dx}$ in terms of $x$ and $y$ and verify that $P$ is a stationary point. A curve is defined by the equation $$2y+e^{2x}y^2=x^2+\frac{2}{e}$$
Find an expression for $\frac{dy}{dx}$ in terms of $x$ and $y$
\begin{align}
2y+e^{2x}y^2 & = x^2+\frac{2}{e} \\
2\frac{dy}{dx}+2e^{2x}y^2+2e^{2x}y\frac{dy}{dx} & = 2x \\ 
\frac{dy}{dx}+e^{2x}y^2+e^{2x}y\frac{dy}{dx} & = x \\ 
\frac{dy}{dx}\left(1+e^{2x}y\right) & = x-e^{2x}y^2 \\ 
\frac{dy}{dx} & = \frac{x-e^{2x}y^2}{1+e^{2x}y} \\ 
\end{align}
Verify that $P$ $(1, \frac{1}{e})$ is a stationary point on the curve.
Stationary point when $\frac{dy}{dx}=0$
$$\frac{x-e^{2x}y^2}{1+e^{2x}y}$$
\begin{align}
& = \frac{(1)-e^{2(1)}(\frac{1}{e})^2}{1+e^{2(1)}(\frac{1}{e})} \\ 
& = \frac{1-e^{2}e^{-2}}{1+e^{2}e^{-1}} \\ 
& = \frac{1}{1+e} \\ 
\end{align}
$$\frac{1}{1+e}\neq0$$
Thanks

You have a mistake in the final part because:
$$
e^2e^{-2}=1
$$
so
$$
1-e^2e^{-2}=0
$$
and
$$
\frac{1-e^{2}e^{-2}}{1+e^{2}e^{-1}}=\frac{0}{1+e}=0
$$
