Solve for $x$ first. You can use the quadratic formula.
$$x^3 - \frac1{x^3} = 108 + 76\sqrt2 \implies x^6 - (108+76\sqrt2)x^3 - 1 = 0$$
$$\implies x^3 = \frac{108+76\sqrt2 \pm \sqrt{(108+76\sqrt2)^2 + 4}}2 = 54 + 38\sqrt2 + 3\sqrt{645 + 456 \sqrt2}$$
$$\implies x = \sqrt[3]{54 + 38\sqrt2 + 3\sqrt{645 + 456 \sqrt2}} = \frac{3+2\sqrt2}2 + \frac12 \sqrt{21 + 12\sqrt2}$$
The hardest part is de-nesting the cube root (see below). Otherwise take WA at its word.
It follows that
$$\frac1x = \frac2{3 + 2\sqrt2 + \sqrt{21 + 12\sqrt2}} = -\frac{3+2\sqrt2}2 + \frac12 \sqrt{21 + 12\sqrt2}$$
and you can easily find $x-\frac1x$ from here.
De-nesting the cube root
Observe that
$$645 + 456 \sqrt2 = (21 + 12\sqrt2) (17 + 12\sqrt2) = (21 + 12\sqrt2) (3 + 2\sqrt2)^2 \\
\implies x^3 = 54 + 38\sqrt2 + 9\sqrt{21+12\sqrt2} + 6\sqrt2\sqrt{21+12\sqrt2}$$
Suppose we can decompose the cube root into the form
$$x = a + b \sqrt2 + c \sqrt{21+12\sqrt2}$$
where $a,b,c\in\Bbb Q$.
Taking cubes on both sides and matching up coefficients, we get the system of equations
$$\begin{cases}
a^3 + 6 a b^2 + 63 a c^2 + 72 b c^2 = 54 \\
2b^3 + 3 a^2 b + 36 a c^2 + 63 b c^2 = 38 \\
7c^3 + a^2 c + 2 b^2 c = 3 \\
2c^3 + a b c = 1
\end{cases}$$
Solving the last equation for $c^2 = 3ab-a^2-2b^2$ and substituting that into the first two equations yields yet another system,
$$\begin{cases}
-62a^3 + 117a^2b + 96ab^2 - 144b^3 = 54 \\
-36a^3 + 48a^2b + 117 ab^2 - 124b^3 = 38
\end{cases}$$
Since both polynomials are homogeneous with degree $3$, suppose that $a,b$ are proportional. Let $b=ka$ where $k\in\Bbb Q$, so we have
$$\begin{cases}
a^3(-62 + 117k + 96k^2 - 144k^3) = 54 \\
a^3(-36 + 48k + 117k^2 - 124k^3) = 38
\end{cases}$$
Eliminate $a^3$ by division to get a cubic equation in $k$.
$$\frac{a^3(-62 + 117k + 96k^2 - 144k^3)}{a^3(-36 + 48k + 117k^2 - 124k^3)} = \frac{27}{19} \implies 612k^3 - 1335k^2 + 927k - 206 = 0$$
Now apply the rational root theorem. If you start with the smallest divisors of $612$ and $206$, you'll soon find $k=\frac23$ (and we can show this is the only rational solution for $k$). It follows that
$$b=\frac23a \implies 16a^3 = 54 \implies a = \frac32 \implies b = 1 \implies c = \frac12$$
Best Answer
As pointed out in the comments, you are wrong in 6th line.
$$h=\sqrt{\sqrt[3]{256}-\sqrt[3]{4}}\ne\sqrt{\sqrt[3]{256}}-\sqrt{\sqrt[3]{4}}$$
The correct way to do this, $$h{=\sqrt{\sqrt[3]{256}-\sqrt[3]{4}}\\ =\sqrt{\sqrt[3]{4^3\cdot4}-\sqrt[3]{4}}\\ =\sqrt{4\sqrt[3]{4}-\sqrt[3]{4}}\\ =\sqrt{3\sqrt[3]{4}}\\ =\sqrt{\sqrt[3]{4\cdot3^3}}\\ =\sqrt{\sqrt[3]{108}}\\ =\sqrt[6]{108} \quad \square}$$
Or, simply you can do, $$h^2{=(2\sqrt[3]{2})^2 - (\sqrt[3]{2})^2\\ =\left(2\sqrt[3]{2} + \sqrt[3]{2}\right)(2\sqrt[3]{2} - \sqrt[3]{2})\\ =\left(3\sqrt[3]{2}\right)\cdot(\sqrt[3]{2}) \\=3(\sqrt[3]{2})^2\\}$$
$$\therefore h=\sqrt3\cdot\sqrt[3]{2}=\sqrt[6]{3^3\cdot2^2}=\sqrt[6]{108}$$