Find the coordinates of the stationary points of $e^x +ye^{-x} = 2e^2$.
So I have differentiated this implicitly, which I think is correct but I'm then unsure how I'd actually solve the equation.
$$\frac d {dx}(e^x) + \frac d {dx}(ye^{-x}) = \frac d {dx} (2e^2)$$
$$\frac d {dx}(e^x) = e^x\quad \frac d {dx}(ye^{-x})=e^{-x}\frac {dy} {dx} - ye^{-x} \quad \frac d {dx}(2e^2)$$
Thus $$e^x +e{-x}\frac{dy}{dx} -ye^{-x}=1$$ $$\frac{dy}{dx} = \frac{1-e^{-x} + ye^{-x}}{e^{-x}}$$ At the stationary point $\frac{dy}{dx}=0$ so $$0=1-e^{x} +ye^{-x}$$ I then multiplied through by $e^x$ and rearranged to get $$y = e^{2x} -e^x$$
Where do I go from here, If I've done everything right up to this stage?