e this in parametric form as $\vec r(x) =
      (x,y(x))$. We now have $z=f(x,y)$ and $\vec r(x)=(x,y(x))$.
      Compute both $Df(x,y)$ and $D\vec r(x)$ symbolically. Don't
      use the function $f(x,y)=x^2+3xy-y^3$ until the last
      step.</li>

      <li>Use the chain rule to compute $D(f(\vec
      r(x)))$. What is $dz/dx$ (i.e., $df/dx$)?</li>

      <li>Since $z$ is held constant, we know that $dz/dx=0$. Use
      this fact, together with previous part, to
      explain why $\ds \frac{dy}{dx} = -\f