Slope of curve in $\mathbb{R}^3$ While doing revision, I came across this problem:
The surface given by $z=x^2-y^2$ is cut by the plane given by $y=3x$, producing a curve in the plane. Find the slope of this curve at the point $(1,3,-8)$.
I tried substituting $y=3x$ into $z=x^2-y^2$, yielding $z=-8x^2$. Then, $\frac{dz}{dx}=-16x=-16$.
However the answer is $-8\sqrt{\frac{2}{5}}$.
Thank you very much for any help.
 A: This is a very badly posed question, and does not have an answer. (Read the comments.)
The following is a solution to a rephrased question which can be answered, however.
Question:
The surface given by $z=x^2−y^2$ is cut by the plane given by $y=3x$, producing a curve in the plane. 
Treating the intersection as a curve in the said plane with vertical axis along $(0,0,1)$ and horizontal axis along $(1,3,0)/\sqrt{10}$, find the slope of this curve at the point (1,3,−8).
Solution:
Any point on the plane has Cartesian coordinates in the form $$\frac{a}{\sqrt{10}}\begin{pmatrix} 1\\3 \\0 \end{pmatrix} + b\begin{pmatrix} 0\\0\\1 \end{pmatrix}.$$ Substituting this into $z=x^2−y^2$, we get $b = -4a^2/5$.
So the "slope" at a point on this intersection, with $a$ and $b$ given, is $$\frac{db}{da} = -\frac{8a}{5}.$$
Setting $$\begin{pmatrix}1 \\3\\-8\end{pmatrix} = \frac{a}{\sqrt{10}}\begin{pmatrix} 1\\3 \\0 \end{pmatrix} + b\begin{pmatrix} 0\\0\\1 \end{pmatrix},$$ we get $a = \sqrt{10}$ and so the "slope" at this point is $-8\sqrt{\frac{2}{5}}$.
Solution using grad:
Let $f:=x^2-y^2-z$ and $g:=y-3x$. At the point $(1,3,-8)$, $\nabla f=(2,-6,-1)$ and $\nabla g=(-3,1,0)$. Their cross product, $(1,3,-16)$, is along the tangent direction of the intersecting curve produced by the surface and the plane, at the point $(1,3,-8)$.
Denote the angle between $(1,3,-16)$ and $(1,3,0)$ (i.e. the "horizontal") by $\theta$. Then, using the dot product, $\cos\theta = \sqrt{\frac{5}{133}}$. The "slope" is $$\tan \theta = - \sqrt{\frac{1}{\cos^2 \theta}-1} = -8\sqrt{\frac{2}{5}},$$ where the negative square root is taken because the "vertical" is along $(0,0,1)$.
