Answer
$x+y-2z=0$
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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$f(x, y) =\sqrt{xy}$
$ \begin{array}{lllll}
f_{x}(x, y) & =\sqrt{y}\cdot\dfrac{\partial}{\partial x}[x^{1/2}] & & f_{y}(x, y) & =\sqrt{x}\cdot\dfrac{\partial}{\partial y}[y^{1/2}]\\
& =\dfrac{\sqrt{y}}{2\sqrt{x}} & & & =\dfrac{\sqrt{x}}{2\sqrt{y}}\\
& & & & \\
f_{x}(1,1) & =\dfrac{1}{2} & & f_{y}(1,1) & =\dfrac{1}{2}
\end{array} $
an equation of the tangent plane at $(x_{0}, y_{0}, z_{0})=(1,1,1) $ is
$z -1=f_{x}(1,1)(x-1)+f_{y}(1,1)(y-1)$
$z-1=\displaystyle \frac{1}{2}(x-1)+\frac{1}{2}(y-1) \qquad/\times 2$
$2z -2=x-1+y-1$
$0 =x+y-2z$