Simplify $\sqrt[4]{\frac{162x^6}{16x^4}}$ is $\frac{3\sqrt[4]{2x^2}}{2}$

$$\sqrt[4]{\frac{162x^6}{16x^4}}=\sqrt[4]{\frac{81\cdot2x^2}{16}}=\frac{3\sqrt[4]{2x^2}}{2}.$$

You made an error:

$$\sqrt[4]{16x^4} = 2\vert x\vert$$

because $\sqrt[4]{16x^4} = \sqrt[4]{(2x)^4}$. (Note the absolute value sign since the value returned is positive regardless of whether $x$ itself is positive or negative.) The rest is fine, so from here, you get

$$\frac{3\sqrt[4]{2x^6}}{2\vert x\vert} = \frac{3\sqrt[4]{2x^4x^2}}{2x} = \frac{3\vert x\vert\sqrt[4]{2x^2}}{2\vert x\vert} = \frac{3\sqrt[4]{2x^2}}{2}$$

As shown in the other answer, it is usually better to simplify within the radical so you don’t mess up with absolute values (for even indices).

$$\sqrt[4]{\frac{162x^6}{16x^4}} = \sqrt[4]{\frac{2\cdot3^4x^2}{2^4}} = \frac{3\sqrt[4]{2x^2}}{2}$$