To find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy the given inequality:

$$f(x^2 + y) \ge \left(\frac{1}{x} + 1\right)f(y)$$

for all $x \in \mathbb{R} \setminus \{0\}$ and $y \in \mathbb{R}$, we'll start by analyzing and simplifying the inequality.

### Step 1: Setting $y = 0$

Set $y = 0$ in the inequality, which gives us:

$$f(x^2) \ge \left(\frac{1}{x} + 1\right)f(0)$$

This must hold for every $x \neq 0$.

### Step 2: Exploring the case $f(0) = 0$

Suppose $f(0) = 0$. Substituting into the inequality from step 1, we obtain:

$$f(x^2) \ge 0$$

This implies that $f(x)$ is non-negative for $x \geq 0$.

### Step 3: Using the inequality for general $y$

Rewrite the inequality:

$$f(x^2 + y) \ge \frac{1}{x} f(y) + f(y)$$

For $f(0) = 0$, using the special case $y = 0$ gives us $f(x^2) \ge \frac{1}{x} \times 0 + 0 = 0$, which simplifies verification of certain conditions. This focuses us on the constraint $f(x) \equiv 0$ for either simplifications or further scenarios.

### Step 4: Assume $f \not\equiv 0$

Assume by contradiction that there is some point $y \in \mathbb{R}$ such that $f(y) \neq 0$. Without loss of generality, we consider $f(y) > 0$.

From the initial inequality condition:

$$f(x^2+y) \ge \left(\frac{1}{x} + 1\right)f(y)$$

Since $\frac{1}{x}$ can be chosen arbitrarily large (by choosing small positive $x$), for large values to both sides, it will imply that $f(x^2 + y)$ must grow unbounded, leading to contradictions in finite bounds known properties of rational functions absent specific multiplicity $y$ affecting $x$, and re-affirms affirmation $f(x) = 0$.

### Step 5: Zero Function Consistency Check

Substitute $f(x) = 0$ into the given inequality:

$$f(x^2 + y) = 0 \quad \text{and} \quad \left(\frac{1}{x} + 1\right)f(y) = 0$$

implies:

$$0 \ge 0$$

Thus, the zero function satisfies the inequality.

### Conclusion

The only function satisfying the given inequality for all $x, y \in \mathbb{R}$ is:

$$\boxed{f(x) = 0}$$