Let's assume that $x_i, y_i > 0,$

The formula simplifies to

$$ n\cdot log(\lambda)+n\cdot log(\frac{1}{\lambda})-\lambda\sum x_i-\frac{1}{\lambda}\sum y_j = n\cdot log(\lambda)+n\cdot log(1)-n\cdot log(\lambda) -\lambda\sum x_i-\frac{1}{\lambda}\sum y_j = -\lambda\sum x_i-\frac{1}{\lambda}\sum y_j $$

and the derivative w.r.t. $\lambda$ is $$ -\sum x_i+\lambda^{-2}\sum y_j $$

so if we define $\bar{x} = \sum x_i$ and $\bar{y}=\sum y_j $, then

$$ -\bar{x}+\lambda^{-2}\bar{y} = 0 $$

$\lambda = \sqrt{\frac{\bar{y}}{\bar{x}}}$, since $\lambda$ needs to be positive

Let's assume that $x_i, y_i > 0,$

The formula simplifies to

$$ n\cdot \log(\lambda)+n\cdot \log(\frac{1}{\lambda})-\lambda\sum x_i-\frac{1}{\lambda}\sum y_j \\=n\cdot \log(\lambda)+n\cdot \log(1)-n\cdot \log(\lambda) -\lambda\sum x_i-\frac{1}{\lambda}\sum y_j \\=-\lambda\sum x_i-\frac{1}{\lambda}\sum y_j $$

and the derivative w.r.t. $\lambda$ is $$ -\sum x_i+\lambda^{-2}\sum y_j $$

so if we define $\bar{x} = \sum x_i$ and $\bar{y}=\sum y_j $, then

$$ -\bar{x}+\lambda^{-2}\bar{y} = 0 $$

$\lambda = \sqrt{\frac{\bar{y}}{\bar{x}}}$, since $\lambda$ needs to be positive