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$