Multivariable Chain Rule. Where is my mistake? Say $x=\sqrt{z^2+y^2}$.
For a function $f(y,z)$, we have
$$
\tag{1}
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} \ .
$$
By definition of $x$, we have
\begin{align}
\frac{\partial y}{\partial x} = \frac{x}{\sqrt{x^2 - z^2}} = \frac{x}{y} \ , \tag{2}\\ 
\frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^2 - y^2}} = \frac{x}{z} \ .\tag{3}\\ 
\end{align}
Therefore 
$$
\tag{4}
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}\frac{x}{y} + \frac{\partial f}{\partial z}\frac{x}{z} \ .
$$

Now assume $f(y,z) = e^{\sqrt{y^2 + z^2}}$. We have 
\begin{align}
\frac{\partial f}{\partial y} = e^{\sqrt{y^2 + z^2}} \frac{y}{\sqrt{y^2 + z^2}}  = e^x\frac{y}{x}\ , \tag{5}\\
\frac{\partial f}{\partial z} = e^{\sqrt{y^2 + z^2}} \frac{z}{\sqrt{y^2 + z^2}} = e^x\frac{z}{x}\ . \tag{6}\\
\end{align}
By $(4)$, we have
$$
\tag{7}
\frac{\partial f}{\partial x} = e^x\frac{y}{x}\frac{x}{y} + e^x\frac{x}{z}\frac{z}{x} = 2e^x\ . 
$$
But that can't be right, because $f = e^x$, so it must be 
$$
\tag{8}
\frac{\partial f}{\partial x} = e^x \neq 2e^x \ . 
$$

So where is my mistake?

What you've written really doesn't make sense. You start with a function $f(u)=e^u$ — a function of a single variable. Then you have $u=g(y,z)$. So, letting $F(y,z) = f(g(y,z))$, we have
$$\frac{\partial F}{\partial y} = f'(g(y,z))\frac{\partial g}{\partial y},$$
and similarly for the $z$ partial.
With $f(u) = e^{\sqrt u}$, this gives
$$\frac{\partial F}{\partial y} = \frac1{2\sqrt{y^2+z^2}}e^{\sqrt{y^2+z^2}}\cdot 
2y = \frac y{\sqrt{y^2+z^2}}e^{\sqrt{y^2+z^2}}.$$
