finding the slope of a 3 dimensional curve

The plane $y = −2$ intersects the surface

$$z = x^3 − y \sin(x + y)$$

in a curve.

What is the slope of the tangent line to this intersection curve at the point $(2, −2, 8)$?

I've tried substituting in $y = -2$ and then finding the derivative of that but I don't think it's right but I'm not too sure what else to do.

• What you did is correct, substitute $y=-2$ since that it where the surface will intersect the plane. Then evaluate the derivative at the x and z coordinate provided. – TSF Nov 19 '15 at 19:30
• Sorry @TonyS.F. I don't really understand what you mean by the second part? I differentiate with respect to x then z and then do I substitute in the x and z co-ordinates given? – Luke McNeil Nov 19 '15 at 19:42
• mathforum.org/library/drmath/view/52062.html – TSF Nov 20 '15 at 2:50