$\lambda$ is typed as lambda, $\alpha$ as alpha.

The PDE $$k^2\frac{\partial^2 u}{\partial x^2}=y^3\frac{\partial u}{\partial y}$$ is separable, so we look for solutions of the form $u(x,t) = X(x)Y(y)$. When solving DE in X and Y use the constants a and b for X and c for Y.
The PDE can be rewritten using this solution as (placing constants in the DE for Y) into
= = $-\lambda$

Note: Use the prime notation for derivatives, so the derivative of $X$ is written as $X^\prime$. Do NOT use $X^\prime(x)$

Since these differential equations are independent of each other, they can be separated
DE in X: $= 0$
DE in Y: $= 0$

Now we solve the separate separated ODEs for the different cases in $\lambda$. In each case the general solution in X is written with constants a and b and the general solution in Y is written with constants c and d. Write the functions alphabetically, so that if the solutions involve cos and sin, your answer would be acos(x) + bsin(x).
Case 1: $\lambda = 0$
$X(x) =$
$Y(y) =$
DE in Y
If $\lambda\neq 0$, the differential equation in Y is first order, linear, and more importantly separable. We separate the two sides as
=
Integrating both sides with respect to $y$ (placing the constant of integration c in the right hand side) we get =
Solving for Y, using the funny algebra of constant where $e^c =c$ is just another constant we get
$Y =$

For $\lambda\neq 0$ we get a Sturm-Louiville problem in X which we need to handle two more cases
Case 2: $\lambda = -\alpha^2$
$X(x) =$
Case 3: $\lambda = \alpha^2$
$X(x) =$

Final Solution
Case 1: $\lambda = 0$
$u =$
Case 2: $\lambda = -\alpha^2$
$u =$
Case 3: $\lambda = \alpha^2$
$u =$