Assume U~Uniform (0,1) Find the probability density function of Y when $$Y = \left(\frac{\alpha}{U^{1/\lambda}}\right)$$
My first step is to find the cdf so i can then easily find the pdf by finding the derivative of the cdf, so far I have put it into $$F(y)=P(Y\le y)$$ form to get $$P\left[log(U)\le \lambda*log\left(\frac{\alpha}{y}\right)\right]$$
I don't know how to continue to find the cdf
I think your have a wrong relation sign. We have
$$P\left(Y<y \right)=P\left(\frac{\alpha}{U^{1/ \lambda}}<y \right)=P\left(\alpha<y\cdot U^{1/ \lambda} \right)$$
$$=P\left(\frac{\alpha}{y}<U^{1/ \lambda} \right)=P\left(\left(\frac{\alpha}{y}\right)^{ \lambda}<U \right)$$
Now you can use the converse probability: $P(U> u)=1-P(U\leq u)$, where
$$F_U(u)=\begin{cases}0, \ \text{if} \ u\leq 0 \\ u, \ \text{if} \ u\in (0,1) \\ 1, \ \text{if} \ u\geq 1 \end{cases}$$
$P\left(\left( U>(\frac{\alpha}{y}\right)^{ \lambda} \right)=1-P\left(\left( U<(\frac{\alpha}{y}\right)^{ \lambda} \right)=1-F_U(u)=1-F_U\left(\left(\frac{\alpha}{y}\right)^{ \lambda} \right)=1-\left(\frac{\alpha}{y}\right)^{ \lambda}$
What is left is to find the three intervals, where $F_Y(y)$ is $<0$, between $0$ and $1$, and equal to $1$.