Partial differentiation of implicit function

Question

How do calculate Partial differentiation of implicit function $$f( x+y+z,x^2+y^2+z^2)=0$$

Answer 1

$$f( x+y+z,x^2+y^2+z^2)=0\equiv f(u,v)=0$$where $~u=x+y+z~$ and $~v=x^2+y^2+z^2$.

Now by chain rule $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}$$ $$\implies \frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}+2x\frac{\partial f}{\partial v}$$

$$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}$$ $$\implies \frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}+2y\frac{\partial f}{\partial v}$$

$$\frac{\partial f}{\partial z}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial z}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial z}$$ $$\implies \frac{\partial f}{\partial z}=\frac{\partial f}{\partial u}+2z\frac{\partial f}{\partial v}$$