Finding the slope at a given point on the intersection of a surface and a plane 
The surface with equation $z = x^{3}+xy^{2}$ intersects the plane with equation $2x−2y =
1$ in a curve. What is the slope of that curve when $x = 1$ and $y = -\frac{1}{2}$?

I'm a little confused about how to go about this question, so, could somebody please help me get started, particularly about how to find the intersection in this case. I presume that the intersection equation must have an $x$ and $y$ remaining as we would have to substitute the given values but I'm not sure how to do it. Could someone help me get going? Details on how to find the intersection (curve) and a hint on how to proceed would help greatly! Thanks!
 A: To find the intersection you could substitute your second equation into the first one, since $y=\frac{2x-1}{2}$ we get:
$z=x^3+xy^2 \Rightarrow z=x^3+x\left (\frac{2x-1}{2} \right )^2$
We now have $y$ and $z$ expressed as functions of $x$ which allows us to give a parametrisation for the curve (that is the intersection between the two surfaces):
$$\vec r(t)=(x(t),y(t),z(t))=\left (t,\frac{2t-1}{2},t^3+t\left (\frac{2t-1}{2} \right )^2 \right)$$
Now find the tangent vector to the curve at your specified point.
Remark: I noticed that the $x=1,y=-1/2$ does not verify the plane equation. So this means that you either have a typo ($x=1,y=1/2$ works) or you have two points: first at $x=1$ and then at $y=-1/2$. If it is the latter case then you need to find what value of $x$ gives you $y=-1/2$ from the plane equation and then substitute this into $\vec r\ '(t)$ (with $x=t$, the parameter).
EDIT: If you haven't covered parametrised curves yet there's another (perhaps even simpler) way of doing this.
Let $f(x,y,z)=x^3+xy^2-z$ and $g(x,y,z)=1-2x+2y$, then your surfaces are the level surface $f(x,y,z)=0$ and $g(x,y,z)=0$. 
Since since $f$ class $\mathscr{C}'$ then the gradient of $f$ at point $P_0\in \left \{P:f(P)=0 \right \}$, $\nabla f(P_0)$, is orthogonal to the level surface at the point $P_0$. 
Similarly the gradient of $g$ is orthogonal to the surface level of $g$.
Now the intersection is a curve which is contained in both surfaces, hence it is orthogonal to both $\nabla f(P)$ and $\nabla g(P)$. Therefore the curve is parallel to the vector product $\nabla f(P) \times \nabla g(P)$. We have therefore obtained a vector which is parallel to the curve at the desired point. 
