= Solution
For $u=XY$, the separated equations are
$$
\frac{X'}{X\sin x}=\tan y\frac{Y'}Y=\lambda,
\qquad X'/X=\lambda\sin x,\qquad Y'/Y=\lambda\cot y.
$$
On a region containing the prescribed points, take $0<x<\pi$ and $0<y<\pi/2$. Integration gives
$$
u=C e^{-\lambda\cos x}(\sin y)^\lambda.
$$
The two specified values, verified in the original PDF, give
$$
C2^{-\lambda}e^{-\lambda\sqrt3/2}=\frac1{4e^{\sqrt3}},\qquad
C\left(\frac{\sqrt3}{2}\right)^\lambda e^{-\lambda/2}=\frac3{4e}.
$$
Dividing these relations eliminates $C$:
$$
\left[\sqrt3\,e^{(\sqrt3-1)/2}\right]^\lambda
=3e^{\sqrt3-1}.
$$
The right-hand side is the square of the positive base on the left, which differs from one. Thus $\lambda=2$, and either original condition gives $C=1$. Therefore
$$
\boxed{u(x,y)=e^{-2\cos x}\sin^2y}.
$$
Its <partial derivatives> satisfy $u_x/\sin x=2u$ and $\tan y\,u_y=2u$, verifying the equation wherever its coefficients are defined. Substitution at the two assigned points gives $1/(4e^{\sqrt3})$ and $3/(4e)$ exactly. The formula extends to any region where $\sin x\ne0$ and $\cos y\ne0$.
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