Answer
$\dfrac{\partial f(-1,-1)}{\partial u}|_{x=0,y=0,z=1}=1$
and $\dfrac{\partial f(-1,-1)}{\partial v}|_{x=0,y=0,z=1}=-2$
Work Step by Step
We have $\dfrac{\partial f}{\partial x}=3x^2; \dfrac{\partial f}{\partial y}=z^2; \dfrac{\partial f}{\partial z}=2yz; \dfrac{\partial x}{\partial u}=2u; \dfrac{\partial y}{\partial u}=1$ and $\dfrac{\partial z}{\partial u}=v; \dfrac{\partial x}{\partial v}=1; \dfrac{\partial y}{\partial v}=2v; \dfrac{\partial z}{\partial v}=u$
We apply the chain rule to obtain:
$\dfrac{\partial f}{\partial u}=\dfrac{\partial f}{\partial x} \dfrac{\partial x}{\partial u}+\dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial u}+\dfrac{\partial f}{\partial z} \dfrac{\partial z}{\partial u} \\=6ux^2+z^2+2vyz$
$\dfrac{\partial f}{\partial v}=\dfrac{\partial f}{\partial x} \dfrac{\partial x}{\partial v}+\dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial v}+\dfrac{\partial f}{\partial z} \dfrac{\partial z}{\partial v} \\=3x^2+2vz^2+2uyz$
Now, $\dfrac{\partial f(-1,-1)}{\partial u}|_{x=0,y=0,z=1}=6(-1)(0)+(1)^2+2(-1)(0)(1)=1$
and $\dfrac{\partial f(-1,-1)}{\partial v}|_{x=0,y=0,z=1}=3(0)+2(-1)(1)+2(-1)(0)(1)=-2$
