Question: I can't seem to find a solution to this.

$$\lim\_{x\to0} \left(\frac{(1+2x)^{1/x}}{e^2}\right)^{1/x}$$

i tried to manipulate to apply Lhopitals rule but i can't see to do it

Answer: 
Whenever you see an exponent, use logs.
$$\require{cancel}\begin{align}y&=\left(\frac{(1+2x)^{\frac1{x}}}{e^2}\right)^\frac1{x} \\ \ln y &=\frac1{x} \ln\left(\frac{(1+2x)^{\frac1{x}}}{e^2}\right)=\frac1{x} \left(\underbrace{\ln(1+2x)^{\frac1{x}}}\_{\frac1{x}\ln(1+2x)}- \cancelto{2}{\ln (e^2)}\right)\\&=\frac1{x} \left(\frac1{x}\ln(1+2x)- 2\right)\\&=\frac{\ln(1+2x)-2x}{x^2}\end{align}$$

Use L'Hôspital, twice:

$$\begin{cases}\text{first: }&\frac{\frac2{1+2x}-2}{2x}=\frac{-4x}{2x+4x^2}\\\text{second:}&\frac{-4}{2+8x}\end{cases}$$

Now it's solveable, plug in $0$ for $x$ and you get $\frac{-4}{2}=-2$ Recall that this is equal to $\ln y$, hence:

$$ \ln y=-2 \\ \therefore y = e^{-2}$$
Once you wrote, as Shavar answered, $$\log(y)=\frac{\log(1+2x)-2x}{x^2}$$ you can also use Taylor series around $x=0$. This gives $$\log(1+2x)=2 x-2 x^2+\frac{8 x^3}{3}+O\left(x^4\right)$$ and then $$\log(y)=-2+\frac{8 x}{3}+O\left(x^2\right)$$ which again approximate $y$ as $$y=\frac{1}{e^2}+\frac{8 x}{3 e^2}+O\left(x^2\right)$$