= Solution
The first two modified stages give
$$
\widetilde{\mathbf u}^{n+1}
=(I+\mu A_x)(I-\mu A_y)^{-1}\mathbf u^n.
$$
The correction stage is
$$
(I-\mu A_x)\mathbf u^{n+1}
=\widetilde{\mathbf u}^{n+1}-\mu A_x\mathbf u^n.
$$
Using $A_xA_y=A_yA_x$,
$$
\begin{aligned}
\widetilde{\mathbf u}^{n+1}-\mu A_x\mathbf u^n
&=\left[(I+\mu A_x)(I-\mu A_y)^{-1}-\mu A_x\right]\mathbf u^n\\
&=(I+\mu^2A_xA_y)(I-\mu A_y)^{-1}\mathbf u^n.
\end{aligned}
$$
Therefore
$$
\boxed{
\mathbf u^{n+1}=D\mathbf u^n,
\qquad
D=(I-\mu A_x)^{-1}
(I+\mu^2A_xA_y)
(I-\mu A_y)^{-1}
}.
$$
On the common eigenvector $v^{(p,q)}$, the eigenvalue of $D$ is
$$
\boxed{
d_{pq}
=\frac{1+\mu^2\lambda_p\lambda_q}
{(1-\mu\lambda_p)(1-\mu\lambda_q)}
}.
$$
Set $a=-\mu\lambda_p>0$ and $b=-\mu\lambda_q>0$. Then
$$
d_{pq}=\frac{1+ab}{(1+a)(1+b)}.
$$
Both numerator and denominator are positive, and
$$
(1+a)(1+b)-(1+ab)=a+b>0.
$$
Hence
$$
0<d_{pq}<1
$$
for every mode and every $\mu>0$. Orthogonal diagonalization now gives $\|D\|_2<1$, so the method is stable without a time-step restriction. This is the <unconditionally stable corrected directional diffusion splitting>.
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