B2. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ for which

$$
f(x y)=x f(y)+3 f(x)+3
$$

holds for all $x, y \in \mathbb{R}$.
B2. We substitute $x=0$ into the functional equation to get $f(0)=3 f(0)+3$, from which we calculate $f(0)=-\frac{3}{2}$. Now we substitute $y=0$ into the functional equation to get $f(0)=x f(0)+3 f(x)+3$, from which we express $f(x)=-\frac{1}{3} x f(0)+\frac{1}{3} f(0)-1=\frac{1}{2} x-\frac{3}{2}$. We can verify that the function $f(x)=\frac{1}{2} x-\frac{3}{2}$ indeed satisfies the given functional equation, so this is the only solution.

Writing the functional equation when $x=0$ 1 point Calculation of $f(0)=-\frac{3}{2}$ 1 point

Writing the functional equation when $y=0$ 1 point

Writing the function $f: f(x)=\frac{x}{2}-\frac{3}{2}$ 2 points

Conclusion that this is the only solution 1 point