By using the definition of kronecker delta $\delta_{KL}$, show that $\delta_{KL}$ is a Cartesian tensor, that is $$\delta ' _{MN}=L_{MK}L_{NL} \delta_{KL}$$ under the rotation $X_K=L_{MK}X' _M$.
I don't really know where to start. What is the result differentiated with respect to? Please help.
Consider the inner product in the Cartesian coordinates, $$ (\mathbf X, \mathbf Y) = X_K Y_K \stackrel{1}{=} \delta_{KL} X_K Y_L \stackrel{2}{=} \delta_{KL} (L_{MK} X'_M) \, (L_{NL} Y'_N) = \delta_{KL} L_{MK} \, L_{NL} X'_M Y'_N, \qquad (1) $$ the summations over repeated indices are assumed. $\stackrel{1}{=}$ follows from the definition of Kronecker delta; $\stackrel{2}{=}$ from the coordinate tranformation formula provided by the problem.
Now the inner product is a scalar, hence invariant under rotations. So the above value must be the same as the value computed in the rotated coordinates: $$ (\mathbf X, \mathbf Y) = X'_N Y'_N = \delta'_{MN} X'_M Y'_N. \qquad (2) $$
Equating (1) and (2), we get, $$ \delta_{KL} L_{MK} \, L_{NL} X'_M Y'_N = \delta'_{MN} X'_M Y'_N. $$ This must hold for any vectors $\mathbf X$ and $\mathbf Y$ with arbitrary components $X'_M$ and $Y'_N$, so $$ \delta'_{MN} = \delta_{KL} L_{MK} \, L_{NL}. $$