Form a PDE by eliminating the arbitrary function Form a PDE by eliminating  the arbitrary function
(i) $\displaystyle F(xy+z^2,x+y+z)=0 $
(ii) $\displaystyle F(x+y+z,x^2+y^2+z^2)=0 $
How do I proceed for any of these?

We set $u=xy+z^2,v=x+y+z$, then the operation of $d$ on (1) leads to:
$$dF(u,v)=\frac{\partial F(u,v)}{\partial u}du+\frac{\partial F(u,v)}{\partial v} dv $$
Thus
$$0=dF(u,v)$$ 
$$\implies 0=du=d(xy+z^2)......(3)$$
$$\text {and   }  0=dv=d(x+y+z)......(4)$$
From (3) and (4) we have:
$$xdy+ydx+2zdz=0......(5)$$
$$dx+dy+dz=0......(6)$$
Thus getting rid of $dy$ we obtain:
$$x(-dx-dz)+ydx+2zdz=0 \implies \frac{y-x}{x-2z}dx=dz......(7)$$
Similarly:
$$xdy+y(-dy-dz)+2zdz=0 \implies \frac{x-y}{y-2z}dy=dz......(8)$$
From (7) and (8) we obtain:
$$\frac{dx}{\frac{x-2z}{y-x}}=\frac{dy}{\frac{y-2z}{x-y}}=dz \implies \frac{dx}{x-2z}=\frac{dy}{2z-y}=\frac{dz}{y-x}$$
Finally we obtain the desired PDE for $z(x,y)$:
$$(x-2z)\frac{\partial z}{\partial x}-(y-2z)\frac{\partial z}{\partial y}=y-x$$
The other implicit equation can be treated in a similar fashion.
-mike
