How this circuit with diodes can be solved?
You are always allowed to select a node and call it \$0\:\text{V}\$. So, assuming you ground (make it \$0\:\text{V}\$) the (-) end, as follows:







simulate this circuit &ndash; Schematic created using CircuitLab

Then \$V_X=\eta\,V_T\cdot\ln\left(\frac12\left[1+\sqrt{5}\,\right]\right)\$. It's pretty much the only possible value, unless the (+) voltage at the top is significantly less than \$10\cdot V_T\$. So, for any applied voltage, \$V\ge 10\cdot V_T\$ (about \$250\:\text{mV}\$ or more), you are just solving \$x\$ where \$e^{^\frac{x}{V_T}}=1+e^{^\frac{-x}{V_T}}\$.

You can work this out entirely from the Shockley equation.

$$\begin{align*}
I_{_{D_1}}&amp;=I_{_{D_2}}+I_{_{D_3}}\\\\
-I_{_\text{SAT}}\left(e^{^\frac{-\left(V-V_X\right)}{\eta\,V_T}}-1\right)&amp;=I_{_\text{SAT}}\left(e^{^\frac{V_X}{\eta\,V_T}}-1\right)-I_{_\text{SAT}}\left(e^{^\frac{-V_X}{\eta\,V_T}}-1\right)\\\\
-e^{^\frac{-\left(V-V_X\right)}{\eta\,V_T}}+1&amp;=e^{^\frac{V_X}{\eta\,V_T}}-e^{^\frac{-V_X}{\eta\,V_T}}\\\\
-e^{^\frac{-V}{\eta\,V_T}}e^{^\frac{V_X}{\eta\,V_T}}-e^{^\frac{V_X}{\eta\,V_T}}+e^{^\frac{-V_X}{\eta\,V_T}}+1&amp;=0\\\\
-\left(e^{^\frac{-V}{\eta\,V_T}}+1\right)e^{^\frac{V_X}{\eta\,V_T}}+e^{^\frac{-V_X}{\eta\,V_T}}+1&amp;=0\\\\
-\left(e^{^\frac{-V}{\eta\,V_T}}+1\right)\left[e^{^\frac{V_X}{\eta\,V_T}}\right]^2+e^{^\frac{V_X}{\eta\,V_T}}+1&amp;=0\\\\\hline\\
\text{set }y&amp;=e^{^\frac{V_X}{\eta\,V_T}}\\\\
\text{quadratic solution:}\quad y &amp;=e^{^\frac{V_X}{\eta\,V_T}}=\frac{1\mp\sqrt{4\,e^{^\frac{-V}{\eta\,V_T}}+5}}{2\left(e^{^\frac{-V}{\eta\,V_T}}+1\right)}\\\\\therefore
V_X&amp;=\eta\,V_T\,\ln\left(\frac{1\mp\sqrt{4\,e^{^\frac{-V}{\eta\,V_T}}+5}}{2\left(e^{^\frac{-V}{\eta\,V_T}}+1\right)}\right)\\\\\hline\\\text{for }V\ge 10 \cdot V_T\text{, }&amp;e^{^\frac{-V}{\eta\,V_T}}\to 0\text{, and }V_X\to \eta\,V_T\,\ln\left(\frac{1+\sqrt{5}}{2}\right)
\end{align*}$$

So, using \$V_T=26\:\text{mV}\$, this is \$V_X\approx \eta\cdot 12.51151\:\text{mV}\$. LTspice shows \$\eta=1.752\$ for the 1N4148 model. This computes out to \$V_X\approx 21.9\:\text{mV}\$. LTspice computes \$V_X=21.886\:\text{mV}\$ using a \$+10\:\text{V}\$ rail. Theory gets you close enough.