Please help me with this one, as I can't seem to finalize the calculus here.
Ex: The equation $$ (y+z) \sin z -y(x+y)=0 $$ defines implicitly the function $$z=z(x,y).$$
Show that: $$z \sin{z} \frac{\partial{z}}{\partial{x}} - y^2 \frac{\partial{z}}{\partial{y}}=0 $$
I tried calculating it by finding $\frac{\partial{z}}{\partial{x}}$ and $\frac{\partial{z}}{\partial{y}}$ directly out of the given equation and replacing it, but I don't think that's correct since I haven't gotten anywhere with my results. Looking forward for some help here.
EDIT: These are the values I found:
$$\frac{\partial{z}}{\partial{x}} = - \frac{F_x}{F_z} = \frac{y}{\sin{z}+(y+z) \cos{z} }$$
$$\frac{\partial{z}}{\partial{y}} = - \frac{F_y}{F_z} = - \frac{\sin{z}-x-2y}{\sin{z}+(y+z) \cos{z} }$$
Going from the PDE and applying the method of characteristics leads to the system $$ \frac{dx}{z\sin z}=\frac{dy}{-y^2}=\frac{dz}{0} $$ which leads to $z(s)=c_1=const.$ and $x=c_1\sin c_1\dfrac1y+c_2$. Thus the general solution has the form $$ 0=F(c_1,c_2)=F\left(z,x-\frac{z\sin z}{y}\right). $$ The original equation can be transformed for $y\ne 0$ to $$ 0=\sin z - \left(x-\frac{z\sin z}{y}\right) -y $$ where the first two terms are constant along a characteristic curve, while $y$ is not constant, so that the whole expression can not be a solution of the PDE.
This last result also proposes that a fix of the original equation by one letter to $$(y+z)\sin z -y(x+z)=0$$ would be a solution of the PDE.