$\sqrt[4]{3^{x^2}\sqrt{3^{x - 1}}} = \sqrt[8]{9^{x + 1}}$
$\left[ 3^{x^2} \left( 3^{x - 1} \right)^{1/2} \right]^{1/4} = \left[ 9^{x + 1} \right]^{1/8}$
$\left( 3^{x^2} \right)^{1/4} \left( 3^{x - 1} \right)^{1/8} = \left( 9^{x + 1} \right)^{1/8}$
Raise both sides of the equation by 8
$\left( 3^{x^2} \right)^2 \left( 3^{x - 1} \right) = 3^{2(x + 1)}$
$\left( 3^{2x^2} \right) \left( 3^{x - 1} \right) = 3^{2x + 2}$
$3^{2x^2 + (x - 1)} = 3^{2x + 2}$
$3^{2x^2 + x - 1} = 3^{2x + 2}$
Hence,
$2x^2 + x - 1 = 2x + 2$
$2x^2 - x - 3 = 0$
$(2x - 3)(x + 1) = 0$
$x = \frac{3}{2} ~ \text{ or } ~ -1$
Check the given equation for x = 3/2
$\sqrt[4]{3^{x^2}\sqrt{3^{x - 1}}} = \sqrt[8]{9^{x + 1}}$
$\sqrt[4]{3^{(3/2)^2}\sqrt{3^{3/2 - 1}}} = \sqrt[8]{9^{3/2 + 1}}$
$\sqrt[4]{3^{9/4}\sqrt{3^{1/2}}} = \sqrt[8]{9^{5/2}}$
$\sqrt[4]{(3^{9/4})(3^{1/4})} = \sqrt[8]{3^{2(5/2)}}$
$\sqrt[4]{3^{9/4 + 1/4}} = \sqrt[8]{3^5}$
$\sqrt[4]{3^{5/2}} = \sqrt[8]{3^5}$
$3^{5/8} = 3^{5/8}$ ← okay!
Check the given equation for x = -1
$\sqrt[4]{3^{x^2}\sqrt{3^{x - 1}}} = \sqrt[8]{9^{x + 1}}$
$\sqrt[4]{3^{(-1)^2}\sqrt{3^{-1 - 1}}} = \sqrt[8]{9^{-1 + 1}}$
$\sqrt[4]{3\sqrt{3^{-2}}} = \sqrt[8]{9^0}$
$\sqrt[4]{3(3^{-1})} = \sqrt[8]{1}$
$\sqrt[4]{3^{1 - 1}} = 1$
$\sqrt[4]{3^0} = 1$
$\sqrt[4]{1} = 1$
$1 = 1$ ← okay!
Thus, x = 3/2 and -1 answer
