Answer
$x+4y-z=0$
Work Step by Step
Suppose $f$ has continuous partial derivatives.
An equation of the tangent plane to the surface $z=f(x, y)$ at the point $P(x_{0}, y_{0}, z_{0})$ is
$z-z_{0}=f_{x}(x_{0}, y_{0})(x-x_{0})+f_{y}(x_{0}, y_{0})(y-y_{0})$
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$f(x, y) =xe^{xy}$
$ \begin{array}{lllll}
f_{x}(x, y) & =\frac{\partial}{\partial x}[x]e^{xy}+x\cdot\frac{\partial}{\partial x}[e^{xy}] & & f_{y}(x, y) & =x\cdot\frac{\partial}{\partial y}[e^{xy}]\\
& =e^{xy}+xye^{xy} & & & =x^{2}e^{xy}\\
& & & & \\
f_{x}(2,0) & =1+0 & & f_{y}(2,0) & =4(1)\\
& =1 & & & =4
\end{array} $
an equation of the tangent plane at $(x_{0}, y_{0}, z_{0})=(2,0,2) $ is
$z-2=f_{x}(2,0)(x-2)+f_{y}(2,0)(y-0)$
$z-2=1(x-2)+4(y-0)$
$z-2=x-2+4y$
$0=x+4y-z$
