Solve for the partial derivative of f with rspect to x

\(\displaystyle{{f}_{{x}}{\left({x},{y}\right)}}=-{4}{x}+{1}\)

Now solve for the partial derivative of f with respect to y

\(\displaystyle{{f}_{{y}}{\left({x},{y}\right)}}={6}{y}\)

Evaluate at the point \(\displaystyle{\left({2},-{1}\right)}\)

\(\displaystyle{{f}_{{x}}{\left({2},-{1}\right)}}=-{7}\)

\(\displaystyle{{f}_{{y}}{\left({2},-{1}\right)}}=-{6}\)

Frame the equation of the tangent plane

\(\displaystyle{x}+{3}=-{7}{x}+{14}-{6}{y}-{6}\)

Simplify

\(\displaystyle-{7}{x}-{6}{y}+{5}={z}\)

\(\displaystyle{{f}_{{x}}{\left({x},{y}\right)}}=-{4}{x}+{1}\)

Now solve for the partial derivative of f with respect to y

\(\displaystyle{{f}_{{y}}{\left({x},{y}\right)}}={6}{y}\)

Evaluate at the point \(\displaystyle{\left({2},-{1}\right)}\)

\(\displaystyle{{f}_{{x}}{\left({2},-{1}\right)}}=-{7}\)

\(\displaystyle{{f}_{{y}}{\left({2},-{1}\right)}}=-{6}\)

Frame the equation of the tangent plane

\(\displaystyle{x}+{3}=-{7}{x}+{14}-{6}{y}-{6}\)

Simplify

\(\displaystyle-{7}{x}-{6}{y}+{5}={z}\)