Answer
$$(x, y, z) =\pm\left(\frac{4}{\sqrt{17}}, \frac{9}{\sqrt{17}},-\frac{2}{\sqrt{17}}\right)$$
Work Step by Step
Given $$\frac{x^{2}}{4}+\frac{y^{2}}{9}+z^{2}=1,\ \ \ \ \mathbf{v}=\langle 1,1,-2\rangle$$
Consider $f(x,y,z)= \frac{x^{2}}{4}+\frac{y^{2}}{9}+z^{2}-1$
Since
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle\\
&=\left\langle \frac{x }{2},\frac{2y}{9},2z \right\rangle
\end{align*}
Since
\begin{align*}
\nabla f_{P}&=k \mathbf{v}\\
\left\langle \frac{x }{2},\frac{2y}{9},2z \right\rangle &=k\langle 1,1,-2\rangle
\end{align*}
Then
$$x= 2k,\ \ \ \ \ \ y = \frac{9k}{2},\ \ \ \ \ \ z=-k $$
Hence
\begin{align*}
\frac{x^{2}}{4}+\frac{y^{2}}{9}+z^{2}&=1\\
\frac{(2 k)^{2}}{4}+\frac{(9 k / 2)^{2}}{9}+(-k)^{2}&=1\\
k^{2}+\frac{9}{4} k^{2}+k^{2}&=1
\end{align*}
Then $$k=\pm \frac{2}{\sqrt {17}} $$
Hence
$$(x, y, z) =\pm\left(\frac{4}{\sqrt{17}}, \frac{9}{\sqrt{17}},-\frac{2}{\sqrt{17}}\right)$$
