It's exercise 23.10 from Mathmatical Methods for Physics and Engineering (Riley K.F., Hobson M.P., Bence S.J. 3d Edition).
Show that the equation
$$f(x) = x^{-1/3} + \lambda\int\limits_{0}^{\infty}f(y)\exp(-xy)dy$$
has a solution of the form $Ax^{\alpha} + Bx^{\beta}$. Determine the values $\alpha$ and $\beta$, and show that those of $A$ and $B$ are $$\frac{1}{1-\lambda^2\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)}\hspace{1cm}\text{and}\hspace{1cm}\frac{\lambda\Gamma\left(\frac{2}{3}\right)}{1-\lambda^2\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)},$$ where $\Gamma(z)$ is the gamma fuction.
This is what I did: I wrote the Laplace transform of $f(x)$ as $\bar f(x)$ (I use the same letter for argument), so the original equation takes the form $$ f(x) = x^{-1/3} + \lambda \bar f(x).$$ Then I took the Laplace transform of that equation and got
$$ \bar f(x) = \Gamma\left(\frac{2}{3}\right)x^{-2/3} + \lambda \int\limits_{0}^{\infty} \frac{f(y)}{x + y}dy.$$ Then I substituted $\bar f(x)$ from the first equation and got $$f(x) = x^{-1/3} + \lambda\Gamma\left(\frac{2}{3}\right)x^{-2/3} + \lambda^2\int\limits_{0}^{\infty}\frac{f(y)}{x + y}dy.$$ And then I took the Laplace transform of this equation, so I got $$\bar f(x) = \Gamma\left(\frac{2}{3}\right)x^{-2/3} + \lambda\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{3}\right)x^{-1/3} + \lambda^2\int\limits_{0}^{\infty}f(y)\exp(xy)\Gamma(0,xy)dy.$$ Now I think that I did something wrong. But I can't understand what exactly that is.
Since you have been given an ansatz, you can use it. Put $f(x) = Ax^\alpha + Bx^\beta$ back into the integral equation to get \begin{align*} Ax^\alpha + Bx^\beta = x^{-1/3} + \lambda\left(A\Gamma(\alpha+1)x^{-(\alpha+1)} + B\Gamma(\beta+1)x^{-(\beta+1)}\right) \end{align*}
Since you want this equation to hold for any $x>0$, you would want one of them, say $\alpha$, to be equal to $-1/3$ since we already have an $x^{-1/3}$ in the equation. $\beta$ would then have to be $-2/3$ because an $x^{-2/3}$ is introduced in the equation, and we don't get anything else. So now we can equate the coefficients of $x^{-1/3}$ and $x^{-2/3}$ on either side to get \begin{align*} A &= 1 + \lambda B\Gamma(1/3), \text{and}\\ B &= \lambda A\Gamma(2/3), \end{align*}
which gives \begin{align*} A &= \frac{1}{1-\lambda^2\Gamma(1/3)\Gamma(2/3)},\text{and}\\ B &= \frac{\lambda\Gamma(2/3)}{1-\lambda^2\Gamma(1/3)\Gamma(2/3)}. \end{align*}