Transforming equation in terms of new independent variables. Transform the equations taking $u$ and $v$ for new independent variables and $w$ for a new function:
$$z_{x x}-2 z_{x y}+\left(1+\frac{y}{x}\right) z_{y} y = 0$$
\begin{align}
u&=x\\
v&=x+y\\
w&=x+y+z
\end{align}
In this example I can't understand, how $z$ is depends on $x,y$, or $u$ and $v$. So how should I differentiate it in a right way?
 A: You want to transform an equation that depends on $z$ into an equation that depends on $w$ and you know how $z$ and $w$ relate to each other.
$$z_{x x}-2 z_{x y}+\left(1+\frac{y}{x}\right) z_{y} y = 0$$
If $z = w - x - y$, it is possible to state that $z_{xx} = w_{xx}$, $z_{xy} = w_{xy}$ and $z_{y} = w_{y} - 1$.
$$w_{x x}-2 w_{x y}+\left(1+\frac{y}{x}\right) (w_{y} - 1) y = 0$$
Now we have to focus on changing the variables.
First, solving the system $\{u = x, v = x + y\}$ for $\{x,y\}$ leads to $\{x = u, y = v - u\}$.
Second, following the chain rule, we know that
$$\begin{matrix}
w,w_u,w_v & -\rightarrow          & u & -\rightarrow          & x\\
          & \diagdown \rightarrow & v & -\rightarrow          & x\\
          &                        &   & \diagdown\rightarrow & y\\
\end{matrix}$$
$$\left\{\begin{matrix}
w_{x} = w_{u} u_{x} + w_{v} v_{x} = w_{u} + w_{v}\\
w_{y} = w_{v} v_{y} = w_{v}
\end{matrix}\right.$$
$$\left\{\begin{matrix}
w_{ux} = w_{uu} u_{x} + w_{uv} v_{x} = w_{uu} + w_{uv}\\
w_{uy} = w_{uv} v_{y} = w_{uv}
\end{matrix}\right.$$
$$\left\{\begin{matrix}
w_{vx} = w_{vu} u_{x} + w_{vv} v_{x} = w_{vu} + w_{vv}\\
w_{vy} = w_{vv} v_{y} = w_{vv}
\end{matrix}\right.$$
$$\left\{\begin{matrix}
w_{xx} = w_{ux} + w_{vx} = w_{uu} + w_{uv} + w_{vu} + w_{vv}\\
w_{xy} = w_{uy} + w_{vy} = w_{uv} + w_{vv}
\end{matrix}\right.$$
Finally,
$$w_{x x}-2 w_{x y}+\left(1+\frac{y}{x}\right) (w_{y} - 1) y = 0$$
$$w_{uu} + w_{uv} + w_{vu} + w_{vv}-2w_{uv} -2 w_{vv}+\left(1+\frac{v-u}{u}\right) (w_{v} - 1) (v-u) = 0$$
