1. **State the problem:** We have a function $\int(x)$ such that its derivative satisfies $$\int'(x) = 6\sqrt{2} \sin(x) (f'(x))^2$$ for all real $x$, and the initial condition $$\int\left(\frac{\pi}{4}\right) = -1.$$ We want to find the value of $$\int\left(-\frac{\pi}{4}\right).$$ 2. **Analyze the given expression:** The derivative of $\int(x)$ depends on $\sin(x)$ and $(f'(x))^2$. Since $(f'(x))^2 \geq 0$, the sign of $\int'(x)$ depends on $\sin(x)$. 3. **Use the Fundamental Theorem of Calculus:** To find $\int\left(-\frac{\pi}{4}\right)$, we can write $$\int\left(-\frac{\pi}{4}\right) = \int\left(\frac{\pi}{4}\right) - \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int'(x) \, dx.$$ 4. **Evaluate the integral of the derivative:** Substitute the expression for $\int'(x)$: $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 6\sqrt{2} \sin(x) (f'(x))^2 \, dx.$$ 5. **Note the symmetry:** $\sin(x)$ is an odd function, and $(f'(x))^2$ is even (square of derivative). The product $\sin(x)(f'(x))^2$ is an odd function. 6. **Integral of an odd function over symmetric limits:** For any odd function $g(x)$, $$\int_{-a}^a g(x) \, dx = 0.$$ Since $\sin(x)(f'(x))^2$ is odd, $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 6\sqrt{2} \sin(x) (f'(x))^2 \, dx = 0.$$ 7. **Calculate the value:** Using step 3, $$\int\left(-\frac{\pi}{4}\right) = \int\left(\frac{\pi}{4}\right) - 0 = -1.$$ **Final answer:** $$\boxed{-1}$$