I recognize that this is an implicit function theorem problem, but I'm not sure exactly how to approach it. I can say that because $\frac{\partial f}{\partial z}(1,-1,2)=3(2)^2+6(1)(-1)(2)-5(1)^2(-1)^2=-5\neq0$, $f=z^3+3xyz^2-5x^2y^2z+14$ can be written in the form $z=g(x,y)$ near the point $(1,-1,2)$, but I'm not sure how this helps.
I also need to determine a locally valid approximation of $z(x,y)$. Any assistance with this is greatly appreciated.
Let $$f(x,y,z)=z^3+3xyz^2-5x^2y^2z+14$$ then we have $ f(1,-1,2)=0$ and $\displaystyle \frac{\partial f}{\partial z}(1,-1,2)=-5\neq0$ So we can apply the Implicit function theorem which garanty the existence of $W\subset \mathbb{R}^2$ neighbour of $(1,-1)$ and $V \subset \mathbb{R}$ neighbour of $2$ and a function $g:W\subset \mathbb{R}^2 \rightarrow V \subset \mathbb{R}$ such that $$ f(x,y,g(x,y))=0 \quad \forall (x,y)\in W$$
And to determine a local approximation you can use Euler method when the partial derivative of $g$ on $W$ are given by
$$ \partial_xg(x,y)= - \frac{\partial_xf(x,y,g(x,y))}{\partial_zf(x,y,g(x,y))} $$ $$ \partial_yg(x,y)= - \frac{\partial_yf(x,y,g(x,y))}{\partial_zf(x,y,g(x,y))} $$