Answer
$z =-7x-6y +5$
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
$z-z_{0}=f_{x}(x_{0}, y_{0})(x-x_{0})+f_{y}(x_{0}, y_{0})(y-y_{0})$
----
$f(x, y) =3y^{2}-2x^{2}+x$
$ \begin{array}{lll}
f_{x}(x, y) =-4x+1 & & f_{y}(x, y)=6y\\
& & \\
f_{x}(2, -1)=-7 & & f_{y}(2, -1)=-6
\end{array} $
an equation of the tangent plane at $(x_{0}, y_{0}, z_{0})=(2,-1,-3) $is
$z -(-3)=f_{x}(2, -1)(x-2)+f_{y}(2, -1)[y -(-1)]$
$z +3=-7(x-2)-6(y+1)$
$z +3=-7x+14-6y-6$
$z =-7x-6y +5$
