We need to find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the given functional equation for all $x, y, z \in \mathbb{R}$:

$$f(f(x) + f(y) + f(z)) = f(f(x) - f(y)) + f(2xy + f(z)) + 2f(xz - yz).$$

To solve this, we'll explore potential forms of $f(x)$ and check if they satisfy the equation.

### Step 1: Check Constant Solutions

First, consider the possibility that $f(x)$ is a constant function, say $f(x) = c$ for all $x \in \mathbb{R}$.

Substituting $f(x) = c$ into the equation, we have:
$$f(f(x) + f(y) + f(z)) = f(3c) = c,$$
and
$$f(f(x) - f(y)) + f(2xy + f(z)) + 2f(xz - yz) = c + c + 2c = 4c.$$

For the equality to hold, we must have $4c = c$, which implies $c = 0$. So, $f(x) = 0$ for all $x$ is a solution.

### Step 2: Check Non-constant Polynomial Solutions

Assume that $f(x)$ is a non-constant polynomial. Checking degree 2 polynomials, let $f(x) = x^2$.

Substitute $f(x) = x^2$ into the functional equation:
Left-hand side:
$$f(f(x) + f(y) + f(z)) = f(x^2 + y^2 + z^2) = (x^2 + y^2 + z^2)^2.$$

Right-hand side:
$$f(f(x) - f(y)) + f(2xy + f(z)) + 2f(xz - yz) = f(x^2 - y^2) + f(2xy + z^2) + 2f(xz - yz).$$

Calculate each term:
- $f(x^2 - y^2) = (x^2 - y^2)^2$,
- $f(2xy + z^2) = (2xy + z^2)^2$,
- $2f(xz - yz) = 2((xz - yz)^2) = 2(xz - yz)^2$.

Upon substitution, we need:
$$(x^2 + y^2 + z^2)^2 = (x^2 - y^2)^2 + (2xy + z^2)^2 + 2(xz - yz)^2.$$

This identity holds, showing that $f(x) = x^2$ is a solution.

### Conclusion

The following functions satisfy the given identity for all $x, y, z \in \mathbb{R}$:

$$f(x) = 0 \quad \text{and} \quad f(x) = x^2.$$

Thus, the functions are:
$$\boxed{f(x) = 0 \text{ and } f(x) = x^2}$$

These are the solutions to the functional equation.