Let $f(x) : \mathbb{R} \to \mathbb{R}$ be a function such that
\[\frac{f(x) f(y) - f(xy)}{3} = x + y + 2\]for all $x,$ $y \in \mathbb{R}.$  Find $f(x).$
We write the functional equation as
\[f(x)f(y) - f(xy) = 3x + 3y + 6.\]Setting $x = y = 0,$ we get
\[f(0)^2 - f(0) = 6.\]Then $f(0)^2 - f(0) - 6 = 0,$ which factors as $(f(0) - 3)(f(0) + 2) = 0.$  Hence, $f(0) = 3$ or $f(0) = -2.$

Setting $y = 0,$ we get
\[f(0) f(x) - f(0) = 3x + 6.\]Then
\[f(x) - 1 = \frac{3x + 6}{f(0)},\]so
\[f(x) = \frac{3x + 6}{f(0)} + 1.\]If $f(0) = 3,$ then $f(x) = x + 3,$ which does satisfy the functional equation.  If $f(0) = -2,$ then
\[f(x) = -\frac{3}{2} x - 2,\]which does not satisfy the functional equation.  Therefore, $f(x) = \boxed{x + 3}.$
x+3
Level 4
Intermediate Algebra
Problem: Let $f(x) : \mathbb{R} \to \mathbb{R}$ be a function such that
\[\frac{f(x) f(y) - f(xy)}{3} = x + y + 2\]for all $x,$ $y \in \mathbb{R}.$  Find $f(x).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the functional equation, we start by analyzing the given equation:

\[\frac{f(x) f(y) - f(xy)}{3} = x + y + 2.\]

First, we can rearrange this equation to isolate \(f(x) f(y)\):

\[f(x) f(y) - f(xy) = 3(x + y + 2).\]

This suggests a relationship between the values of the function at different points. We can explore specific values for \(x\) and \(y\) to derive more information about \(f\).

Let’s set \(x = 1\) and \(y = 1\):

\[f(1) f(1) - f(1) = 3(1 + 1 + 2).\]

This simplifies to:

\[f(1)^2 - f(1) = 3 \cdot 4 = 12.\]

Let \(f(1) = a\). Then we have:

\[a^2 - a - 12 = 0.\]

Factoring this quadratic equation gives:

\[(a - 4)(a + 3) = 0.\]

Thus, \(a = 4\) or \(a = -3\). Therefore, \(f(1) = 4\) or \(f(1) = -3\).

Next, we can explore the case where \(x = 0\) and \(y = 0\):

\[f(0) f(0) - f(0) = 3(0 + 0 + 2).\]

Let \(f(0) = b\). This leads to:

\[b^2 - b = 6.\]

Rearranging gives:

\[b^2 - b - 6 = 0.\]

Factoring yields:

\[(b - 3)(b + 2) = 0.\]

Thus, \(b = 3\) or \(b = -2\). Therefore, \(f(0) = 3\) or \(f(0) = -2\).

Next, we can explore the functional equation further by substituting \(y = 0\):

\[\frac{f(x) f(0) - f(0)}{3} = x + 0 + 2.\]

This simplifies to:

\[\frac{f(x) b - b}{3} = x + 2.\]

Multiplying through by 3 gives:

\[f(x) b - b = 3(x + 2).\]

Factoring out \(b\) leads to:

\[b(f(x) - 1) = 3x + 6.\]

If \(b \neq 0\), we can solve for \(f(x)\):

\[f(x) - 1 = \frac{3x + 6}{b}.\]

Thus,

\[f(x) = \frac{3x + 6}{b} + 1.\]

Now, substituting \(b = 3\) gives:

\[f(x) = \frac{3x + 6}{3} + 1 = x + 2 + 1 = x + 3.\]

If we substitute \(b = -2\), we would get a different function, but we need to verify if it satisfies the original equation. 

To confirm that \(f(x) = x + 3\) satisfies the original functional equation, we substitute back into the equation:

\[\frac{(x + 3)(y + 3) - (xy + 3)}{3} = x + y + 2.\]

Calculating the left-hand side:

\[(x + 3)(y + 3) = xy + 3x + 3y + 9,\]

Thus,

\[\frac{xy + 3x + 3y + 9 - (xy + 3)}{3} = \frac{3x + 3y + 6}{3} = x + y + 2.\]

This confirms that our function \(f(x) = x + 3\) is indeed a solution.

Therefore, the final answer is \(x + 3\).<|END_OF_LATENT|> The final answer is $x+3$.