The problem in question is
$(1+x)y''(x)+\frac12y'(x) + \lambda y(x)=0$
subject to $y(0)=y(3)=0$.
Thanks to a suggested change of variable ($y(x)=u(\sqrt{1+x})$), I've managed to find the general solution to the equation: $y(x)=c_1\cos(2\sqrt{\lambda}\sqrt{1+x})+c_ 2\sin(2\sqrt{\lambda}\sqrt{1+x})$ for $c_1, c_2 \in \mathbb{R}$.
However, the boundary values don't help much to find the problem's solution (is it even possible?), so I don't know how to proceed in finding the eigenvalues and eigenfunctions. Does anyone have any suggestion?
Edit: this sums up to finding $\lambda$ such that the system $Sc = 0$ below admits solutions other than $c=0$:
$$Sc=0\Leftrightarrow\begin{bmatrix}\cos(2\sqrt{\lambda}) & \sin(2\sqrt{\lambda}) \\ \cos(4\sqrt{\lambda}) & \sin(4\sqrt{\lambda})\end{bmatrix}\begin{bmatrix}c_1 \\ c_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$
Taking the determinant of $S$, we get $\cos(2\sqrt{\lambda})\sin(4\sqrt{\lambda}) - \cos(4\sqrt{\lambda})\sin(2\sqrt{\lambda})=\sin(2\sqrt{\lambda})$. If $Sc=0$ for $c\neq0$, then $\det(S)=0$, so we're pretty much done:
$$\sin(2\sqrt{\lambda})=0\Leftrightarrow\lambda=\frac{k^2\pi^2}4,\, k\in\mathbb{Z}.$$
After this, we can find that $c_1=0$ and $c_2$ can be anything, so the solution to this problem would be $y(x) = c_2\sin(k\pi\sqrt{1+x})$.
You get by introducing a specific phase constant into the solution formula that the arguments of the sine and cosine are zero at $x=0$ and thus the solution satisfying the left boundary condition as $$ y(x)=c\sin(2\sqrtλ(\sqrt{1+x}-1)). $$ Then for the right boundary condition you need $$ 0=y(3)=c\sin(2\sqrtλ)\implies 2\sqrtλ=k\pi,~~k\in\Bbb N_{>0}. $$
Why the phase term: The function class $y(x)=c_1\cos(ϕ(x))+c_2\sin(ϕ(x))$ is equal by basic trigonometric identities to the class $y(x)=d_1\cos(ϕ(x)+A)+d_2\sin(ϕ(x)+A)$. Now set $A=-ϕ(0)$ to get that $y(0)=0$ implies $d_1=0$, so that $y(x)=d_2\sin(ϕ(x)-ϕ(0))$.