The value of _ x → 0 ((x^2 ^2 x)/(x^2 - ^2 x)) is:

Let $L = \lim_{x \to 0} \dfrac{x^2\sin^2 x}{x^2 - \sin^2 x}$

Dividing the numerator and the denominator by $x^4$, we get:

$L = \lim_{x \to 0} \dfrac{\left(\dfrac{\sin x}{x}\right)^2}{\dfrac{x^2 - \sin^2 x}{x^4}}$

Factorizing the denominator:

$L = \lim_{x \to 0} \dfrac{\left(\dfrac{\sin x}{x}\right)^2}{\left(\dfrac{x - \sin x}{x^3}\right) \left(\dfrac{x + \sin x}{x}\right)}$

Using the standard limits:

$\lim_{x \to 0} \dfrac{\sin x}{x} = 1$

$\lim_{x \to 0} \dfrac{x - \sin x}{x^3} = \dfrac{1}{6}$ (using Taylor series expansion $\sin x = x - \dfrac{x^3}{3!} + \dots$)

$\lim_{x \to 0} \dfrac{x + \sin x}{x} = \lim_{x \to 0} \left(1 + \dfrac{\sin x}{x}\right) = 1 + 1 = 2$

Substituting these limits into the expression:

$L = \dfrac{1^2}{\left(\dfrac{1}{6}\right) \times 2} = \dfrac{1}{\dfrac{1}{3}} = 3$

Answer: $3$