Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$
So
$$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$
Then
$$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$
That's
$$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}(1+\sqrt[3]{4-3x})}$$
Finally
$$\frac{3}{(1+\sqrt[3]{4-3x})}$$
But this evaluates to
$$\frac{3}{2}$$
When the answer should be
$$1$$
Where did I fail?

If you let $t=4-3x,\;$ so $x=\frac{1}{3}(4-t)$,  you get
$\displaystyle\lim_{x\to 1}\frac{1-\sqrt[3]{4-3x}}{x-1}=\lim_{t\to1}\frac{1-\sqrt[3]{t}}{\frac{1}{3}(1-t)}=\lim_{t\to1}\frac{3(1-t^{1/3})}{(1-t^{1/3})(1+t^{1/3}+t^{2/3})}=\lim_{t\to1}\frac{3}{1+t^{1/3}+t^{2/3}}=1$
