$2 x^2-8 y^2=x^2+x y^2+100$
$x^2-8 y^2=x y^2+100$
$2 x-16 y \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}+y^2$
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 x-y^2}{2 x y+16 y}$
Suppose $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 x-y^2}{2 x y+16 y}=0$
$2 x=y^2 $
$\therefore \frac{x^2-8 x}{x^2+2 x^2+100}=\frac{1}{2} $
$2 x^2-16 x=3 x^2+100 $
$x^2+16 x+100=0 $
$(x+8)^2+36=0$
No solution since $(x+8)^2+36>0$ for $x \in \mathbb{R}$.
[Or using discriminant: Since $16^2-4(1)(100)=-144<0$, there’s, no real roots.]
$\therefore$ There is no stationary point since $\frac{\mathrm{d} y}{\mathrm{~d} x} \neq 0$ for $x \in \mathbb{R}$.
