Let's determine those positive integers for which

$$
\frac{3}{\sqrt{x}}+\frac{2}{\sqrt{y}}=\frac{1}{\sqrt{2}}
$$
Solution. Express $y$ from the equation:

$$
\begin{gathered}
\frac{3}{\sqrt{x}}+\frac{2}{\sqrt{y}}=\frac{1}{\sqrt{2}}, \quad \frac{2}{\sqrt{y}}=\frac{\sqrt{x}-3 \sqrt{2}}{\sqrt{2 x}} \\
\sqrt{y}=\frac{2 \sqrt{2 x}}{\sqrt{x}-3 \sqrt{2}}, \quad y=\left(\frac{2 \sqrt{2 x}}{\sqrt{x}-3 \sqrt{2}}\right)^{2}=\frac{8 x}{x+18-6 \sqrt{2 x}}
\end{gathered}
$$

Since $y$ is an integer, the right side must also be an integer. Since $x$ is also an integer, the fraction can only be an integer if $\sqrt{2 x}$ is rational, which is true if it is also an integer, so $2 x$ is a perfect square. Based on this, $x=2 \cdot k^{2}$, where $k$ is a positive integer. Similarly,

$$
x=\left(\frac{3 \sqrt{2 y}}{\sqrt{y}-2 \sqrt{2}}\right)^{2}=\frac{18 y}{y+8-4 \sqrt{2 y}}
$$

so $2 y$ is also a perfect square. $y=2 \cdot m^{2}$, where $m$ is a positive integer. Substituting the above into the original equation and multiplying by $\sqrt{2}$, we get a Diophantine equation:

$$
\frac{3}{k}+\frac{2}{m}=1
$$

After multiplying and factoring:

$$
\begin{gathered}
3 m+2 k=k \cdot m, \quad k \cdot m-3 m-2 k=0 \\
(k-3) m-2(k-3)=6, \quad(k-3)(m-2)=6
\end{gathered}
$$

The positive integer solutions of the equation are obtained from the divisor pairs:

| $k-3$ | $m-2$ | $k$ | $m$ | $x$ | $y$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 6 | 4 | 8 | 32 | 128 |
| 2 | 3 | 5 | 5 | 50 | 50 |
| 3 | 2 | 6 | 4 | 72 | 32 |
| 6 | 1 | 9 | 3 | 162 | 18 |