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I am studying up for my final however our prof didn't give us solutions for the review he gave us. I am a bit confused with this question and if anyone could shed some light on it I would appreciate it.

Question: Describe all points in space where the gradient vector $f(x,y,z) = xy+z^2$ is parallel to the vector $<2,-3,1>$.

This is what i was thinking... I found the gradient and if the gradient has to be parallel to the vector $<2,-3,1>$ could i compute the cross porduct $\nabla f \ \times <2,-3,1> =0$ and solve for $x, y$ and $z$ or is there something else I am missing. Again any help is appreciated.

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Yes, your suggestion to calculate the cross product will work. (Check your solution against this answer:…) – in_wolframAlpha_we_trust Apr 26 '13 at 7:36

Your gradient $\nabla f =\; <y,x,2z>$ must satisfy $\nabla f = k <2,-3,1>$ with some real constant $k$. I.e. $x$, $y$, and $z$ must satisfy

$$\frac{y}{2} = -\frac{x}{3} = 2z$$

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