Multivariable Chain Rule Doubt Im learning Partial differential Equations now ( introductory course ). While in calculus course I have learnt that ,

If $f=f(x,y,z)$ where $x,y,z$ are independent variables then,
$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz$$ in case $f=f(x,y,z)$ and $x=x(u,v)$, $y=y(u,v)$, $z=z(u,v)$ where $u,v$ are independent variables.
then, $$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial u}$$ but in case of PDE formulation of the eq. $$f(x,y,z,a,b)=0\tag{$\circledast$}$$ where $a,b$ are arbitrary constant
How can I differentiate $\circledast$ w.r.t. $x$ because here $z$ is not independent variable but $z=z(x,y)$.

Im aware of the proof for multivariable chain rule in case of 2 independent variables variables I know same can be extrapolated to 3 independent variables
 A: Firstly, recall implicit differentiation from single variable calculus. If you have an equation like $x^3+y^2-64=0$, then you can think of $y$ as an implicit function of $x$, and differentiate both sides with respect to $x$ using the chain rule: $3x^2+2y\dfrac{\mathrm dy}{\mathrm dx}=0$ so that $\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2}{2y}$, which holds whether or not $y=\sqrt{64-x^3}$ or $y=-\sqrt{64-x^3}$ (as long as $y\ne 0$).
Your situation is the same thing, just with more variables. If $z$ is an implicit function of $x$ and $y$, then we can use the chain rule to write $0=\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial x}+\dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial x}+\dfrac{\partial f}{\partial a}\dfrac{\partial a}{\partial x}+\dfrac{\partial f}{\partial b}\dfrac{\partial b}{\partial x}$. But $\dfrac{\partial a}{\partial x}$ and $\dfrac{\partial b}{\partial x}$ are $0$ because $a$ and $b$ are constants, and $\dfrac{\partial y}{\partial x}=0$ since $x$ and $y$ are our independent variables. So this reduces to  $0=\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial z}\dfrac{\partial z}{\partial x}$ so that $\dfrac{\partial z}{\partial x}=\dfrac{-\partial f/\partial x}{\partial f/\partial z}$.
