How to prove ?

Let z=f(x,y) and $z {e}^{z} - {x}^{2} y = 0$. Prove that $x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y} = 1 - \frac{1}{1 + z} , \mathmr{if} z \ne 1$.

Jun 24, 2018

$z {e}^{z} - {x}^{2} y = 0 q \quad \star$

Differentials:

$\mathrm{dz} \setminus {e}^{z} + z {e}^{z} \setminus \mathrm{dz} - 2 x y \setminus \mathrm{dx} - {x}^{2} \mathrm{dy} = 0$

$\boldsymbol{{e}^{z} \left(1 + z\right) \setminus \mathrm{dz} - 2 x y \setminus \mathrm{dx} - {x}^{2} \mathrm{dy} = 0}$

For the partials:

• $\mathrm{dy} = 0$

${z}_{x} = \frac{2 x y}{{e}^{z} \left(1 + z\right)}$

• $\mathrm{dx} = 0$

${z}_{y} = {x}^{2} / \left({e}^{z} \left(1 + z\right)\right)$

$\therefore x {z}_{x} - y {z}_{y} = \frac{2 {x}^{2} y - {x}^{2} y}{{e}^{z} \left(1 + z\right)} = \frac{{x}^{2} y}{{e}^{z} \left(1 + z\right)}$

From $\star , q \quad = \frac{z {e}^{z}}{{e}^{z} \left(1 + z\right)}$

$= \frac{1 + z - 1}{\left(1 + z\right)} = 1 - \frac{1}{1 + z}$

Can see the case for $z \ne \text{-1}$ but not for $z \ne \text{1}$