$a_{ij}$ is a rotation matrix that satisfies $\hat{e}'_i=a_{ij}\hat{e}_j$. Show that $\epsilon_{lmn}a_{mi}a_{nj}=\epsilon_{ijk}a_{lk}.$
Using the result from above, how can I show that $\vec{x}\times \vec{y}$ transforms like a rank$-1$ tensor under the rotation $a_{ij}?$
I am struggling understanding tensors.
So, I know that right-handed Cartesian coordinate system is defined by unit vectors that satisfy $\hat{e}_i \times\ \hat{e}_j = \epsilon_{ijk}\hat{e}_k$ and that $\epsilon$ is the Levi-Civita symbol. But I don't know how to solve this problem.
You mention "the tensor notation of a determinant" in the comments. I don't know quite how your definitions have been given to you, but I imagine that you already know something like this: $$\epsilon_{pmn}a_{pk}a_{mi}a_{nj}=\epsilon_{kij}\operatorname{det}(a).$$ Then in the case $a$ is a rotation matrix, we know $\operatorname{det}(a)=1$, so we have $$\epsilon_{pmn}a_{pk}a_{mi}a_{nj}=\epsilon_{kij}.$$ We also know that $a$'s inverse is its transpose, so that $a_{pk}a_{lk}=\delta_{pl}$. So if we multiply both sides of our equation by $a_{lk}$ we get $$\epsilon_{pmn}\delta_{pl}a_{mi}a_{nj}=\epsilon_{kij}a_{lk}$$ and hence $$\epsilon_{lmn}a_{mi}a_{nj}=\epsilon_{ijk}a_{lk}.$$ For the second bit of the question, start from the fact that $(\vec{x}\times\vec{y})_i=\epsilon_{ijk}x_jy_k$.