Catenoid is a minimal surface i want to show that the catenoid is a minimal surface. I have given 

$f:I \times (0,2\pi)\longrightarrow \mathbb{R}^3$ with $f(r,\phi)=\left( \begin{array}{c}\cosh(r) \;\cos(\phi)\\\cosh(r) \;\sin(\phi)\\r\end{array} \right)$.

I know that:

$f$ is minimal surface $\Longleftrightarrow$ $\Delta f=0$.

$f$ is given in polar coordinates so i have to calculate the following: 

$\Delta f= \frac{\partial^2f}{\partial r^2}+\frac{1}{r}\frac{\partial f}{\partial r}+\frac{1}{r^2}\frac{\partial^2f}{\partial \phi^2}$

$\frac{\partial f}{\partial r}= \left( \begin{array}{c}\sinh(r) \;\cos(\phi)\\\sinh(r) \;\sin(\phi)\\1\end{array} \right)$ , $\frac{\partial^2f}{\partial r^2}=\left( \begin{array}{c}\cosh(r) \;\cos(\phi)\\\cosh(r) \;\sin(\phi)\\0\end{array} \right)$

$\frac{\partial f}{\partial \phi}=\left( \begin{array}{c}-\cosh(r) \;\sin(\phi)\\\cosh(r) \;\cos(\phi)\\0\end{array} \right)$ ,  $\frac{\partial^2 f}{\partial \phi^2}=\left( \begin{array}{c}-\cosh(r) \;\cos(\phi)\\-\cosh(r) \;\sin(\phi)\\0\end{array} \right)$.

But when I put all together I can not show that $\Delta f$ is 0. What did I do wrong?
Can someone help me please?

Thanks in advance.
I think you are a little bit confused about the harmonic characterization of (conformally immersed) minimal surfaces.  

We know that every regular $2$-dimensional surface can be described locally in isothermal coordinates (i.e. for neighborhood of the surface, there's a coordinate map that preserves angles aka is conformal aka has 1st fundamental form satisfying $E=G$, $F=0$).  

So we can cover the surface by a family of coordinate maps $$\vec{x_\alpha}(u,v) = \big(x_1(u,v), x_2(u,v), x_3(u,v)\big)$$

with each $\vec{x_\alpha}$ conformally mapping an open subset of $\mathbb{R}^2$ to $\mathbb{R}^3$.

The harmonic characterization says that the surface is minimal iff for each $\vec{x}_\alpha$ in such a family, the coordinates $x_i(u,v)$ are harmonic functions with respect to the coordinates (u,v).  I would advise going back to look at the proof of this characterization for clarification, and thinking about geometrically what it means to be conformal (preserve angles).  

I think where you have been misled is in thinking of this as a polar parametrization and using the so-called "polar form of the laplacian." 

In your case you have a conformal coordinate map describing the entire catenoid in coordinates $r, \varphi$.  What does $$\big(\cosh(r)\cos(\varphi)\big)_{rr} + \cosh(r)\cos(\varphi)\big)_{\varphi \varphi}$$ look like?
This not a complete answer but that's too long for me to post it in comment.


Preliminary for differential geometry of surfaces

\begin{align*}
  \mathbf{x}(u,v)
  &amp;= \begin{pmatrix} x(u,v) \\ y(u,v) \\ z(u,v) \end{pmatrix} \\
  \mathbf{x}_u &amp;= \frac{\partial \mathbf{x}}{\partial u} \\
  \mathbf{x}_v &amp;= \frac{\partial \mathbf{x}}{\partial v} \\
  \mathbf{N} &amp;=
  \frac{\mathbf{x}_u \times \mathbf{x}_v}{|\mathbf{x}_u \times \mathbf{x}_v|}
  \tag{unit normal vector} \\
\end{align*}

First fundamental form
$$\mathbb{I}=
\begin{pmatrix} E &amp; F \\ F &amp; G \end{pmatrix}= 
\begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}
\begin{pmatrix} \mathbf{x}_{u} &amp; \mathbf{x}_{v} \end{pmatrix} $$

Second fundamental form
$$\mathbb{II}=
 \begin{pmatrix} e &amp; f \\ f &amp; g \end{pmatrix}=
-\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}
 \begin{pmatrix} \mathbf{x}_{u} &amp; \mathbf{x}_{v} \end{pmatrix}$$

Metric
$$ds^2=E\, du^2+2F\, du\, dv+G\, dv^2$$

Element of area
$$dA=|\det \mathbb{I}| \, du \, dv
=|\mathbf{x}_u \times \mathbf{x}_v| \, du\, dv
=\sqrt{EG-F^2} \, du\, dv$$

Principal curvatures

Let $\begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}=
\mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}$ 
where $\mathbb{A}=
\begin{pmatrix} a_{11} &amp; a_{12} \\ a_{21} &amp; a_{22} \end{pmatrix}$.

Now
\begin{align*}
   \begin{pmatrix} \mathbf{N}_{u} \\ \mathbf{N}_{v} \end{pmatrix}
   \begin{pmatrix} \mathbf{x}_{u} &amp; \mathbf{x}_{v} \end{pmatrix} &amp;=
   \mathbb{A} \begin{pmatrix} \mathbf{x}_{u} \\ \mathbf{x}_{v} \end{pmatrix}
   \begin{pmatrix} \mathbf{x}_{u} &amp; \mathbf{x}_{v} \end{pmatrix} \\
  -\begin{pmatrix} e &amp; f \\ f &amp; g \end{pmatrix} &amp;=
   \mathbb{A} \begin{pmatrix} E &amp; F \\ F &amp; G \end{pmatrix} \\
   \mathbb{A} &amp;=
  -\begin{pmatrix} e &amp; f \\ f &amp; g \end{pmatrix}
   \begin{pmatrix} E &amp; F \\ F &amp; G \end{pmatrix}^{-1}
\end{align*}

The principal curvatures $k_{1}, k_{2}$ are the eigenvalues of $-\mathbb{A}$.

Mean curvature

$$H=\frac{k_{1}+k_{2}}{2}
=-\frac{1}{2} \operatorname{tr} \mathbb{A}
=\frac{eg-f^2}{EG-F^2}$$

Gaussian curvature

$$K=k_{1} k_{2}
=(-1)^{2} \det \mathbb{A}
=\frac{eG-2fF+gE}{2(EG-F^2)}$$


  For minimal surface, $$H=0$$