\begin{align} f'(x) &= f(x)^2 + f^{-1}(x) + \int_{x}^{-\infty} \frac{e^t}{t} \, dt \end{align}
How to find $f(x)$
What i do so far
\begin{align} f'(x) &= f(x)^2 + f^{-1}(x) + \int_{x}^{-\infty} \frac{e^t}{t} , dt \end{align}
\begin{align} (f^{-1})'(x) &= \frac{1}{f'(f^{-1}(x))} \end{align}
\begin{align} f'(x) &= f(x)^2 + \frac{1}{f'(f^{-1}(x))} + \int_{x}^{-\infty} \frac{e^t}{t} , dt \end{align}
u = f(x): \begin{align} u &= f(x) \ f'(x) &= \frac{du}{dx} \end{align}
\begin{align} \frac{du}{dx} &= u^2 + \frac{1}{f'\left(f^{-1}(u)\right)} + \int_{x}^{-\infty} \frac{e^t}{t} , dt \end{align}
\begin{align}\frac{du}{dx} &= u^2 + \frac{1}{f'\left(f^{-1}(u)\right)} - \frac{e^x}{x} \end{align}
\begin{align} \textbf{1. Original Differential Equation:} \nonumber \\ f'(x) &= (f(x))^2 + f^{-1}(x) + \text{Ei}(x) \\ \textbf{2. Assume a Power Series Representation for } f(x): \nonumber \\ f(x) &= \sum_{n=0}^{\infty} a_n x^n \\ \textbf{3. Substitute into the Differential Equation:} \nonumber \\ \sum_{n=0}^{\infty} a_n n x^{n-1} &= \left(\sum_{n=0}^{\infty} a_n x^n\right)^2 + \frac{1}{\sum_{n=0}^{\infty} a_n x^n} + \text{Ei}(x) \\ \textbf{4. Derive the Recurrence Relation for } a_n: \nonumber \\ n a_n &= \sum_{k=0}^{n} a_k a_{n-k} + \frac{1}{a_0} \delta_{n,0} \\ \textbf{5. Define the Generating Function } A(x): \nonumber \\ A(x) &= \sum_{n=0}^{\infty} a_n x^n \\ \textbf{6. Substitute for } f(x) \textbf{ and Derive the Quadratic Equation for } A(x): \nonumber \\ (A(x))^2 - \frac{1}{x} A(x) - \frac{1}{a_0 x} &= 0 \\ \textbf{7. Solve the Quadratic Equation for } A(x): \nonumber \\ A(x) &= \frac{1}{2x} \left(1 \pm \sqrt{1 + \frac{4}{a_0 x}}\right) \\ \textbf{8. Simplify the Series for } A(x): \nonumber \\ A(x) &= \frac{1}{2x} \left(1 \pm \sum_{k=0}^{\infty} \binom{1/2}{k} \left(\frac{4}{a_0 x}\right)^k\right) \\ \textbf{9. Match Coefficients and Express in Terms of } _2F_1: \nonumber \\ A(x) &= \frac{1}{2x} \sum_{k=0}^{\infty} \binom{1/2}{k} \left(\frac{4}{a_0}\right)^k x^{-k} \\ &= \frac{1}{2x} \, _2F_1\left(1, \frac{1}{2}; 2; \frac{-4}{a_0 x}\right) \\ \textbf{10. Express } a_0 \textbf{ in terms of } a_1 \textbf{ and } a_2: \nonumber \\ a_0 &= a_2 - \frac{a_1^2}{2} \\ \textbf{11. Express } f(x) \textbf{ in terms of Gauss Hypergeometric Function:} \nonumber \\ f(x) &= \left(\frac{2}{a_2 - \frac{a_1^2}{2}}\right) \, _2F_1\left(1, \frac{1}{2}; 2; \frac{-4x}{a_2 - \frac{a_1^2}{2}}\right) \end{align}