To solve the problem, we need to determine all polynomials $P(x)$ with real coefficients satisfying the equation:

$$P(x)^2 + P\left(\frac{1}{x}\right)^2 = P(x^2)P\left(\frac{1}{x^2}\right)$$

for all $x$.

### Step 1: Analyze the Equation

Let's start by inspecting the given functional equation. Set $x = 1$:

$$P(1)^2 + P(1)^2 = P(1)P(1) \implies 2P(1)^2 = P(1)^2$$

This implies either $P(1) = 0$ or $P(1) = \text{undefined}$. The latter does not apply here, so let us assume $P(1) = 0$.

### Step 2: Consider Special Values

Next, substitute $x = 0$:

$$P(0)^2 + P\left(\frac{1}{0}\right)^2 \text{ is undefined as } P\left(\frac{1}{0}\right) \text{ is undefined.}$$

This prompts that the function might inherently contain no constant non-zero term, as imaginary or undefined inputs do not yield a valid expression.

### Step 3: Assume $P(x) = 0$ and Check

Suppose $P(x) = 0$. Substituting into the original equation gives:

$$0^2 + 0^2 = 0 \cdot 0,$$

which simplifies to $0 = 0$, thus satisfying the equation trivially for all $x$.

### Step 4: Check for Non-trivial Solutions

Consider whether there could be a non-zero polynomial satisfying the given condition.

1. Assume $P(x) = c$ where $c \neq 0$. Substituting back, we get:

$$c^2 + c^2 = c \cdot c \implies 2c^2 = c^2,$$

which fails unless $c = 0$. Therefore, $c \neq 0$ gives no valid solution.

2. Suppose $P(x)$ is of degree $n$. Then each side of the equation must be a polynomial of degree $2n$. Moreover, due to symmetry in substitution $x$ and $\frac{1}{x}$, and enforcing both degrees equal, $P(x)$ cannot maintain a balance without nullifying effectively.

Thus, the only consistent polynomial across scenarios that satisfy the functional equation is the zero polynomial.

Therefore, the polynomial $P(x)$ satisfying the original condition is:
$$\boxed{0}$$