Answer
The point on the graph of $z = x{y^3} + 8{y^{ - 1}}$ where the tangent plane is parallel to $2x + 7y + 2z = 0$ is
$\left( {\frac{3}{2}, - 1, - \frac{{19}}{2}} \right)$
Work Step by Step
Write $f\left( {x,y} \right) = x{y^3} + 8{y^{ - 1}}$. So, the partial derivatives are
${f_x} = {y^3}$, ${\ \ \ }$ ${f_y} = 3x{y^2} - 8{y^{ - 2}}$
Let $\left( {a,b} \right)$ be the point on the graph of $f$. By Eq. (2) the tangent plane is given by
$z = f\left( {a,b} \right) + {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_y}\left( {a,b} \right)\left( {y - b} \right)$
$z = a{b^3} + 8{b^{ - 1}} + {b^3}\left( {x - a} \right) + \left( {3a{b^2} - 8{b^{ - 2}}} \right)\left( {y - b} \right)$
$z = a{b^3} + 8{b^{ - 1}} + {b^3}x - a{b^3} + \left( {3a{b^2} - 8{b^{ - 2}}} \right)y - 3a{b^3} + 8{b^{ - 1}}$
$z = {b^3}x + \left( {3a{b^2} - 8{b^{ - 2}}} \right)y - 3a{b^3} + 16{b^{ - 1}}$
Or
${b^3}x + \left( {3a{b^2} - 8{b^{ - 2}}} \right)y - z = 3a{b^3} - 16{b^{ - 1}}$
According to Eq. (5) of Section 13.5, the normal vector to this plane is
${{\bf{n}}_1} = \left( {{b^3},3a{b^2} - 8{b^{ - 2}}, - 1} \right)$
It is desired that the tangent plane be parallel to $2x + 7y + 2z = 0$, which normal vector is ${\bf{n}} = \left( {2,7,2} \right)$. Thus, the normal vectors ${\bf{n}} = \left( {2,7,2} \right)$ and ${{\bf{n}}_1} = \left( {{b^3},3a{b^2} - 8{b^{ - 2}}, - 1} \right)$ are parallel. So, we can write
$\frac{{{b^3}}}{2} = \frac{{3a{b^2} - 8{b^{ - 2}}}}{7} = \frac{{ - 1}}{2}$
From $\frac{{{b^3}}}{2} = \frac{{ - 1}}{2}$ we obtain $b = - 1$.
Substituting $b = - 1$ in $\frac{{3a{b^2} - 8{b^{ - 2}}}}{7} = \frac{{ - 1}}{2}$ gives $a = \frac{3}{2}$.
So, the point on the graph of $z = x{y^3} + 8{y^{ - 1}}$ where the tangent plane is parallel to $2x + 7y + 2z = 0$ is
$\left( {a,b,f\left( {a,b} \right)} \right) = \left( {\frac{3}{2}, - 1,f\left( {\frac{{ - 1}}{2},\frac{3}{2}} \right)} \right) = \left( {\frac{3}{2}, - 1, - \frac{{19}}{2}} \right)$
