# How to deal with the total differential of implicit function equation

I want to find the total differential of $$z=z(x, y)$$, $$z=z(x, y)$$ satisfies the implicit function equation $$(x+1) z-y^{2}=x^{2} f(x-z, y)$$ (function $$f(u, v)$$ is differentiable).

Dt[(x + 1) z[x, y] - y^2 == x^2*f[x - z[x, y], y], z[x, y]] 

But the above result is not in the form of $$\mathrm{d} z=p(\mathrm{x}, \mathrm{y},\mathrm{z(x,y)}) \mathrm{d} \mathrm{x}+\mathrm{q}(\mathrm{x}, \mathrm{y},\mathrm{z(x,y)}) \mathrm{d} \mathrm{y}$$.

What should I do to get the form I want?

Test examples:

$$\left.\boldsymbol{d} z\right|_{(0,1)}=-\boldsymbol{d} x+2 \boldsymbol{d} y$$