$ F(x^2+y^2+z^2, z^2-2xy)=0 $
I checked some online examples and I started by letting
$u=x^2+y^2+z^2$, $v=z^2-2xy$
$\frac{\partial u}{\partial x}=2x+2zp$
$\frac{\partial u}{\partial y}=2y+2zq$
$\frac{\partial v}{\partial x}=z^2-2xy$
$\frac{\partial v}{\partial y}=2zq-2x$
Finding Jacobian
$0= 4(x+zp)(zq-x)-4(zp-y)(y+zq)$
But I am not sure of this method and it does not give the answer.