Answer
$$z=-14+5 x+10 y$$
Work Step by Step
Given $$f(x, y)= x^{2}y+xy^3 \quad \text { at } \quad(2,1)$$
Since
\begin{aligned}
f_{x}(x, y) &=2xy+y^3 \\
f_{x}(2,1) &= 5
\end{aligned}
and
\begin{aligned}
f_{y}(x, y) &=x^2+3xy^2 \\
f_{y}(2,1) &= 10
\end{aligned}
Then the tangent plane given by
\begin{aligned}
z&=f(2,1)+f_{x}(2,1)(x-2)+f_{y}(2,1)(y-1) \\
&=6+5(x-2)+10(y-1) \\
&=-14+5 x+10 y
\end{aligned}
