How I can solve this circuit with diodes?
You know the Shockley diode equation:

$$I_\text{D}=I_\text{SAT}\left(e^{^\frac{V_\text{D}}{\eta\,V_T}}-1\right)$$

Since the emission coefficient, \$\eta\$, isn't specified we can assume it is \$\eta=1\$ and simplify a bit (or you can keep it in, if you want.) Here, I'll ignore it.

The only voltage you need to compute is the common node shared by all three diodes (unlabeled in your diagram.) Let's call it \$V_\text{X}\$. Then nodal analysis says that the sum of the currents into the node must equal the currents out of the node. So it's easy to set up the following (since \$I_{\text{SAT}_1}=I_{\text{SAT}_2}\$, I'm just replacing any instance of \$I_{\text{SAT}_2}\$ with \$I_{\text{SAT}_1}\$ below):

$$\begin{align*}
I_{\text{SAT}_1}\left(e^{^\frac{V_\text{X}}{V_T}}-1\right)+I_{\text{SAT}_3}\left(e^{^\frac{-V_\text{X}}{V_T}}-1\right)&amp;=I_{\text{SAT}_1}\left(e^{^\frac{65\:\text{mV}-V_\text{X}}{V_T}}-1\right)\\\\
e^{^\frac{V_\text{X}}{V_T}}+\frac{I_{\text{SAT}_3}}{I_{\text{SAT}_1}}\cdot e^{^\frac{-V_\text{X}}{V_T}}-\frac{I_{\text{SAT}_3}}{I_{\text{SAT}_1}}&amp;=e^{^\frac{65\:\text{mV}}{V_T}}\cdot e^{^\frac{-V_\text{X}}{V_T}}\\\\
\left(e^{^\frac{V_\text{X}}{V_T}}\right)^2-\frac{I_{\text{SAT}_3}}{I_{\text{SAT}_1}}\cdot e^{^\frac{V_\text{X}}{V_T}}+\frac{I_{\text{SAT}_3}}{I_{\text{SAT}_1}}&amp;=e^{^\frac{65\:\text{mV}}{V_T}}
\end{align*}$$

Set \$y=e^{^\frac{V_\text{X}}{V_T}}\$ and the above becomes this quadratic:

$$y^2+\left(-\frac{I_{\text{SAT}_3}}{I_{\text{SAT}_1}}\right)y+\left(\frac{I_{\text{SAT}_3}}{I_{\text{SAT}_1}}-e^{^\frac{65\:\text{mV}}{V_T}}\right)=0$$

Solve that for \$y\$ (pick the reasonable one of the two answers) and then find \$V_\text{X}=V_T\,\operatorname{ln}\:y\$. (I think in your situation \$V_T\approx 25\:\text{mV}\$, but feel free to use whatever you think is appropriate there.)