I got this limit:
$$\lim_{x\to1}\frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1} \implies \lim_{x\to1}\frac{\frac{x-1}{\sqrt[3]{x²}+\sqrt[3]{x}+1}}{\sqrt[4]{x}-1} \implies \lim_{x\to1}\frac{x-1}{\sqrt[3]{x²}+\sqrt[3]{x}+1}*\frac{1}{\sqrt[4]{x}-1}*\frac{\sqrt[4]{x}+1}{\sqrt[4]{x}+1} \implies \lim_{x\to1}\frac{\sqrt[4]{x}+1}{\sqrt[3]{x²}+\sqrt[3]{x}+1} \xrightarrow[x\to 1]{}\frac{2}{3}.$$ But in the book the answer is $$ \frac{4}{3}.$$ I cannot find the mistake in my calculation.
You are cancelling $x-1$ in the numerator with $(\sqrt[4]{x}-1)(\sqrt[4]{x}+1) = \sqrt x - 1$ in the denominator.
But l'Hôpital's rule would have been easier!
If fractional exponents are difficult to work with, you can also try a variable substitution: let $x = u^{12}$, so that $\sqrt[3]{x} = u^4$ and $\sqrt[4]{x} = u^3$. Then we have $$\frac{\sqrt[3]{x} - 1}{\sqrt[4]{x} - 1} = \frac{u^4 - 1}{u^3 - 1} = \frac{(u-1)(u+1)(u^2+1)}{(u-1)(u^2+u+1)} = \frac{(u+1)(u^2+1)}{u^2+u+1}.$$ And since $x \to 1$ implies $u \to 1$, the limit is simply $$\frac{(1+1)(1^2+1)}{1^2+1+1} = \frac{4}{3},$$ as claimed.
That's probably because $\dfrac{x-1}{\sqrt[4]x-1}=\sqrt[4]{x^3}+\sqrt[4]{x^2}+\sqrt[4]{x}+1$.
Your calculation mistake is already pointed out by TonyK.
I don't really understand what prevents you from using the standard limit: $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ This is in fact the least used of all standard limits (most used one is probably $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$). We have to only notice that here $a = 1$ and then $$\lim_{x \to 1}\frac{\sqrt[3]{x} - 1}{\sqrt[4]{x} - 1} = \lim_{x \to 1}\dfrac{\dfrac{x^{1/3} - 1}{x - 1}}{\dfrac{x^{1/4} - 1}{x - 1}} = \frac{1/3}{1/4} = \frac{4}{3}$$
