How to find the limit of this function using L'Hospital's rule? 
$$\lim_{x\rightarrow 0} \,\,\left( \sqrt[3]{1+2x+x^3} - \frac{2x}{2x+3}   \right) ^ {\frac1{x^3}} $$

I have already tried several options, but the only answer I have gotten so far is
$e^{\infty}$, which is incorrect. The correct answer is $e^\frac{43}{81}$, which is easy to get by using Taylor series, but our task was to get the same one by using L'Hospital's rule. Can you help me with it?

The standard approach here is to take logarithm and then evaluate the limit. If $f(x) $ denotes the expression under limit and it tends to a limit $L$ then
\begin{align}
\log L&=\lim_{x\to 0}\frac{1}{x^3}\log\left(\sqrt[3]{1+2x+x^3}-\frac{2x}{2x+3}\right)\notag\\
&=\lim_{x\to 0}\frac{g(x)}{x^3}\cdot\frac{\log(1+g(x))}{g(x)}\notag\\
&=\lim_{x\to 0} \frac{g(x)} {x^3}\notag\\
&=\lim_{x\to 0}\frac{1}{x^3}\left(\sqrt[3]{1+2x+x^3}-1-\frac{2x}{2x+3}\right)\notag
\end{align}
The algebraic limit at the end can be evaluated using standard techniques. Here is one approach which requires some algebra. If $$A=\sqrt[3]{1+2x+x^3},B=\frac{4x+3}{2x+3}$$ then $A,B$ both tend to $1$ and $$A^3-B^3=x^3+\frac{(1+2x)(2x+3)^3-(4x+3)^3}{(2x+3)^3}=x^3+\frac{16x^3(1+x)} {(2x+3)^3}$$ so that $(A^3-B^3)/x^3\to 43/27$. And therefore $$\frac{A-B} {x^3}=\frac{A^3-B^3}{x^3}\cdot\frac{1}{A^2+AB+B^2}\to \frac{43}{81}$$ Thus the desired limit is $L=e^{43/81}$.
Advanced tools like L'Hospital's Rule and Taylor series are often unnecessary for simple limit problems and instead a little bit of algebra helps.

Incidentally this exercise helps us to know that $(3+2x)/(3+x)$ is a very good approximation to $\sqrt[3]{1+x}$ for small values of $x$. For example if $x=0.1$  then we have $$\sqrt[3]{1.1}=1.032280\dots,\frac{3.2}{3.1}=1.032258\dots$$
