We are tasked with finding all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all real numbers $x$ and $y$, the following functional equation holds:

$$f(x)f(y) + f(x+y) = xy.$$

Step 1: Plug-in Specific Values

First, let's test for simple values of $x$ and $y$.

1. Set $y = 0$:
$$f(x)f(0) + f(x+0) = x \cdot 0 \Rightarrow f(x)f(0) + f(x) = 0 \Rightarrow f(x)(f(0) + 1) = 0.$$
Since this must hold for all $x$, we have two possibilities:

- $f(x) = 0$ for all $x$, which is easily seen not to satisfy the equation for, say, $x = y = 1$.
- $f(0) = -1$.

**Step 2: Determine $f(x)$ Using Another Strategy**

Now let's use the fact that $f(0) = -1$.

2. Set $x = y$:
$$f(x)f(x) + f(2x) = x^2 \Rightarrow f(x)^2 + f(2x) = x^2.$$
From this equation, we can solve for $f(2x)$:
$$f(2x) = x^2 - f(x)^2.$$

Step 3: Consider Another Combination

Consider setting $x = 1$ and $y = 1$:
$$f(1)f(1) + f(2) = 1 \Rightarrow f(1)^2 + f(2) = 1.$$

Also note from setting $x = 1$ and previously finding $f(0) = -1$:
$$f(1)f(0) + f(1) = 0 \Rightarrow -f(1) + f(1) = 0 \Rightarrow f(1) = 0.$$

Using $f(1) = 0$ in the above, $f(1)^2 + f(2) = 1$ becomes:
$$0^2 + f(2) = 1 \Rightarrow f(2) = 1.$$

Step 4: Propose and Verify a General Form

Given our earlier results and a few trials, let us propose $f(x) = x - 1$ as one solution:

1. Calculate $f(x)f(y) + f(x+y)$ using $f(x) = x-1$:

$$(x-1)(y-1) + (x+y-1) = (xy - x - y + 1) + (x + y - 1) = xy.$$

The calculation confirms that $f(x) = x - 1$ satisfies the functional equation.

Additionally, propose $f(x) = -x - 1$ as another solution and check:

2. Calculate $f(x)f(y) + f(x+y)$ using $f(x) = -x-1$:

$$(-x-1)(-y-1) + (-(x+y)-1) = (xy + x + y + 1) - (x + y + 1) = xy.$$

Thus, $f(x) = -x - 1$ also satisfies the functional equation.

Conclusion:

The solutions to the functional equation are:
$$\boxed{f(x) = x - 1 \text{ or } f(x) = -x - 1}.$$

Both forms satisfy the original equation for all real numbers $x$ and $y$.