Finding $f\in\mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$ and $\deg(f)\leq 3$. What's wrong with my approach?

Let $e = g + \sqrt{3}h$, $f = j + \sqrt{3}k$. $$\begin{eqnarray}(x^2 -2)(ex+f) &=& a(x+\sqrt{3})^3 +b(x+\sqrt{3})^2 +c(x+\sqrt{3}) +d -x \\ (g + \sqrt{3}h)x^3 + (j + \sqrt{3}k)x^2 - 2(g + \sqrt{3}h)x - 2(j + \sqrt{3}k) &=& a(x^3 + 3\sqrt{3}x^2 + 9x + 3\sqrt{3}) +b(x^2 + 2\sqrt{3}x + 3) + cx + \sqrt{3}c +d - x \\ (g + \sqrt{3}h)x^3 + (j + \sqrt{3}k)x^2 - 2(g + \sqrt{3}h)x - 2(j + \sqrt{3}k) &=& ax^3 + (b + 3\sqrt{3}a)x^2 + (9a + c - 1 + 2\sqrt{3}b)x + (3b + d + 3\sqrt{3}a + \sqrt{3}c) \\ \end{eqnarray}$$

So $$\begin{eqnarray}g &=& a \\ h &=& 0 \\ j &=& b \\ k &=& 3a \\ -2g &=& 9a+c-1 \\ -2h &=& 2b \\ -2j &=& 3b + d \\ -2k &=& 3a + c \end{eqnarray}$$

Quickly reduces to $$\begin{eqnarray}h = b = j = d &=& 0 \\ g &=& a \\ k &=& 3a \\ c &=& -9a \\ 2a &=& 1 \\ \end{eqnarray}$$

So there's no flaw in your reasoning: the flaw is in the part you didn't include in the question.