Express $z$ in terms of $x$ and $y$, i.e., find $z= f(x,y)$ I've been banging my head against the wall for a while now:
$x = s^2 - t^2$
$y = s + t$
$z = s^2 + 3t$
Express $z$ in terms of $x$ and $y$.

$$x=s^{2}-t^{2}=(s+t)(s-t)$$
so
$$s+t=\frac{x}{s-t}$$
$$s-t=\frac{x}{s+t}=\frac{x}{y}$$
$$(s+t) + (s-t) = 2s=\frac{x}{s-t}+\frac{x}{y}$$
$y=s+t$, so $t=y-s$ and therefore:
$$2s=\frac{x}{2s-y}+\frac{x}{y}=x(\frac{1}{2s-y}+\frac{1}{y})$$
$$2s=x(\frac{y}{2sy-y^2}+\frac{2s-y}{2sy-y^2})=x(\frac{2s}{2sy-y^2})$$
$$1=\frac{x}{2sy-y^2}$$
$$2sy-y^2=x$$
$$2sy=x+y^2$$
$$s=\frac{x+y^2}{2y}$$
From $z=s^2+3t$ we have:
$$z=(\frac{x+y^2}{2y})^{2}+3t$$
$y=s+t$ so $t=y-\frac{x+y^2}{2y}$ and finally:
$$z=(\frac{x+y^2}{2y})^{2}+3(y-\frac{x+y^2}{2y})$$
Pretty sure this is correct... 
