To solve the inequality \( \frac{2 + x}{x^2} \neq 0 \), we need to determine the values of \( x \) for which the expression \( \frac{2 + x}{x^2} \) is not equal to zero.

First, let's rewrite the inequality:

\[ \frac{2 + x}{x^2} \neq 0 \]

For a fraction to be non-zero, its numerator must be non-zero (since the denominator cannot be zero in any case, as division by zero is undefined). Therefore, we need to find when the numerator \( 2 + x \) is not equal to zero.

1. **Find when the numerator is zero:**

\[ 2 + x = 0 \]

Solving for \( x \):

\[ x = -2 \]

So, the numerator \( 2 + x \) is zero when \( x = -2 \). For the fraction to be non-zero, \( x \) must not be \(-2\).

2. **Consider the denominator:**

The denominator \( x^2 \) must not be zero, which means \( x \neq 0 \).

3. **Combine the conditions:**

The expression \( \frac{2 + x}{x^2} \) is non-zero if:

\[ x \neq -2 \]
\[ x \neq 0 \]

Therefore, the solution to the inequality \( \frac{2 + x}{x^2} \neq 0 \) is:

\[ x \in \mathbb{R} \setminus \{-2, 0\} \]

In interval notation, this can be written as:

\[ (-\infty, -2) \cup (-2, 0) \cup (0, \infty) \]

This means \( x \) can be any real number except \(-2\) and \( 0 \).