Where is the mistake in this limit calculation? 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.

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 $$\lim_{x \to 1}\frac{\sqrt[3]{x} - 1}{\sqrt[4]{x} - 1} = \lim_{x \to 1}\dfrac{\dfrac{x^{1/3} - 1}{x - 1}}{\dfrac{x^{1/4} - 1}{x - 1}} = \frac{1/3}{1/4} = \frac{4}{3}$$
