$f:\mathbb{R}\rightarrow \mathbb{R}$ satisfies the equation
\[f(x)f(y)-af(xy)=x+y\]
, for every real numbers $x,y$. Find all possible real values of $a$.
1. **Substitute \( x = 0 \) and \( y = 0 \) into the given functional equation:**
   \[
   f(0)f(0) - af(0) = 0 + 0 \implies f(0)^2 - af(0) = 0
   \]
   This can be factored as:
   \[
   f(0)(f(0) - a) = 0
   \]
   Therefore, we have two possible solutions:
   \[
   f(0) = 0 \quad \text{or} \quad f(0) = a
   \]

2. **Consider the case \( f(0) = 0 \):**
   Substitute \( y = 0 \) into the original equation:
   \[
   f(x)f(0) - af(0) = x + 0 \implies f(x) \cdot 0 - a \cdot 0 = x \implies 0 = x
   \]
   This is a contradiction for any \( x \neq 0 \). Therefore, \( f(0) = 0 \) is not a valid solution.

3. **Consider the case \( f(0) = a \):**
   Substitute \( y = 0 \) into the original equation:
   \[
   f(x)f(0) - af(0) = x + 0 \implies f(x)a - a^2 = x \implies af(x) - a^2 = x
   \]
   Solving for \( f(x) \):
   \[
   af(x) = x + a^2 \implies f(x) = \frac{x + a^2}{a} = a + \frac{x}{a}
   \]

4. **Verify the solution \( f(x) = a + \frac{x}{a} \) in the original equation:**
   Substitute \( f(x) = a + \frac{x}{a} \) and \( f(y) = a + \frac{y}{a} \) into the original equation:
   \[
   f(x)f(y) - af(xy) = (a + \frac{x}{a})(a + \frac{y}{a}) - a\left(a + \frac{xy}{a}\right)
   \]
   Simplify the left-hand side:
   \[
   (a + \frac{x}{a})(a + \frac{y}{a}) = a^2 + a\frac{y}{a} + a\frac{x}{a} + \frac{xy}{a^2} = a^2 + x + y + \frac{xy}{a^2}
   \]
   Simplify the right-hand side:
   \[
   a\left(a + \frac{xy}{a}\right) = a^2 + xy
   \]
   Therefore:
   \[
   a^2 + x + y + \frac{xy}{a^2} - (a^2 + xy) = x + y
   \]
   Simplify:
   \[
   a^2 + x + y + \frac{xy}{a^2} - a^2 - xy = x + y \implies x + y = x + y
   \]
   This is true for all \( x \) and \( y \).

5. **Determine the possible values of \( a \):**
   Substitute \( x = a \) and \( y = a \) into the original equation:
   \[
   f(a)f(a) - af(a^2) = a + a \implies (a + \frac{a}{a})^2 - a\left(a + \frac{a^2}{a}\right) = 2a
   \]
   Simplify:
   \[
   (a + 1)^2 - a(a + a) = 2a \implies (a + 1)^2 - 2a^2 = 2a
   \]
   Expand and simplify:
   \[
   a^2 + 2a + 1 - 2a^2 = 2a \implies -a^2 + 2a + 1 = 2a \implies -a^2 + 1 = 0 \implies a^2 = 1
   \]
   Therefore:
   \[
   a = 1 \quad \text{or} \quad a = -1
   \]

6. **Verify the functions for \( a = 1 \) and \( a = -1 \):**
   - For \( a = 1 \):
     \[
     f(x) = 1 + x
     \]
     Substitute into the original equation:
     \[
     (1 + x)(1 + y) - 1(1 + xy) = x + y \implies 1 + x + y + xy - 1 - xy = x + y
     \]
     This is true.

   - For \( a = -1 \):
     \[
     f(x) = -1 - x
     \]
     Substitute into the original equation:
     \[
     (-1 - x)(-1 - y) - (-1)(-1 - xy) = x + y \implies 1 + x + y + xy - 1 - xy = x + y
     \]
     This is true.

The final answer is \( \boxed{a = 1, -1} \)