Answer
$$z=-34-20 x+16 y$$
Work Step by Step
Given $$f(x, y)=2 x^{2}-4 x y^{2} \quad \text { at }(-1,2)$$
Since
\begin{aligned}
f_{x}(x, y) &=4 x-4 y^{2} \\
f_{x}(-1,2) &=-20
\end{aligned}
and
\begin{aligned}
f_{y}(x, y) &=-8xy \\
f_{y}(-1,2) &=16
\end{aligned}
Then the tangent plane is given by
\begin{align*}
z &=f(-1,2)+f_{x}(-1,2)(x+1)+f_{y}(-1,2)(y-2) \\
&=18-20(x+1)+16(y-2) \\
&=18-20 x-20+16 y-32 \\
&=-34-20 x+16 y
\end{align*}
