Finding derivative of $\sqrt[3]{x}$ using only limits

Here is a hint: Use the identity $(a^3-b^3)=(a-b)\cdot(a^2+ab+b^2)$ with $a$, $b$ being suitable cube roots. Otherwise, the method is similar to the one you tried.

$$\lim_{h \to 0} \frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h} $$ $$=\lim_{h \to 0} \frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h} \cdot \frac{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}}{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}} $$

$$=\lim_{h \to 0} \frac{x+h-x}{h((x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3})}$$ $$=\lim_{h \to 0} \frac{1}{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}}$$ $$=\frac{1}{(x)^{2/3} + x^{1/3}(x)^{1/3} + (x)^{2/3}}$$ $$=\frac{1}{3x^{2/3}}$$ $$=\frac{x^{-2/3}}{3}$$ As obtained from the $Dx^{n} = n.x^{n-1}$

I understand that the point of this exercise is to apply the limit definition of the derivative to a function where the limit calculation is "tricky". But it's worth noting that if $F(x,y)=0$ identically (as in $y-\sqrt[3]{x}=0$ in this problem) then $\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$.

So given that $x=y^3$, we have that $\frac{dx}{dy}=3y^2$ (either using the power rule or a simpler limit computation). That makes $\frac{dy}{dx}=\frac{1}{3y^2}=\frac{1}{3(\sqrt[3]{x})^2}=\frac{1}{3}x^{-\frac{2}{3}}$.