The gradient is given by \(\displaystyle▽{f}={\left(\begin{array}{c} {d}\frac{{f}}{{\left.{d}{x}\right.}}\\{d}\frac{{f}}{{\left.{d}{y}\right.}}\\{d}\frac{{f}}{{\left.{d}{z}\right.}}\end{array}\right)}\)

Now, \(\displaystyle{\left({d}\frac{{f}}{{\left.{d}{x}\right.}}\right)}=\frac{{d}}{{\left.{d}{x}\right.}}{\left({x}{y}-{4}{z}{e}+{35}\right)}=\frac{{d}}{{\left.{d}{x}\right.}}{\left({x}{y}\right)}+\frac{{d}}{{\left.{d}{x}\right.}}{\left(-{4}{z}{e}+{35}\right)}={y}\) (when taking the partial derivative with respect to x, all other yariables are treating as constants). Similarly,

\(\displaystyle{d}\frac{{f}}{{\left.{d}{y}\right.}}=\frac{{d}}{{\left.{d}{y}\right.}}{\left({x}{y}\right)}+\frac{{d}}{{\left.{d}{y}\right.}}{\left(-{4}{z}{e}+{35}\right)}={x}\) and \(\displaystyle{d}\frac{{f}}{{\left.{d}{z}\right.}}=\frac{{d}}{{\left.{d}{z}\right.}}{\left(-{4}{z}{e}\right)}+\frac{{d}}{{\left.{d}{z}\right.}}{\left({x}{y}+{35}\right)}=-{4}{e}\) Therefore, \(\displaystyle▽{f}={\left({y},{x},-{4}{e}\right)}\) To find the gradient at(3,-3,2), plug in \(x=3, y=-3, z=2\) into(1): \(\displaystyle▽{f{{\left({3},-{3},{2}\right)}}}={\left(-{3},{3},-{4}{e}\right)}\)