To solve the equation \(\left| x \left| x^2 + \frac{3}{4} \right| \right| = x\), we need to consider the properties of absolute values and analyze different cases based on the value of \(x\).

### Case 1: \(x \geq 0\)
When \(x\) is non-negative, \(|x| = x\). The equation simplifies to:
\[ x \left| x^2 + \frac{3}{4} \right| = x \]

Since \(x \neq 0\) (otherwise the equation would be trivially true), we can divide both sides by \(x\):
\[ \left| x^2 + \frac{3}{4} \right| = 1 \]

This gives us two sub-cases to consider:
1. \( x^2 + \frac{3}{4} = 1 \)
2. \( x^2 + \frac{3}{4} = -1 \) (which is not possible since the left side is always non-negative)

For the first sub-case:
\[ x^2 + \frac{3}{4} = 1 \]
\[ x^2 = 1 - \frac{3}{4} \]
\[ x^2 = \frac{1}{4} \]
\[ x = \pm \frac{1}{2} \]

Since we are considering \(x \geq 0\), we have:
\[ x = \frac{1}{2} \]

### Case 2: \(x < 0\)
When \(x\) is negative, \(|x| = -x\). The equation simplifies to:
\[ -x \left| x^2 + \frac{3}{4} \right| = x \]

Since \(x \neq 0\), we can divide both sides by \(x\):
\[ - \left| x^2 + \frac{3}{4} \right| = 1 \]

This equation is not possible because the left side is always non-positive while the right side is positive.

### Conclusion
The only solution to the equation is:
\[ x = \frac{1}{2} \]