The given equation is equivalent to $$y\ln{x} = x\ln{y}\implies \frac{\ln{x}}{x}=\frac{\ln{y}}{y}$$

Taking derivative on both sides gives $$\begin{align*}\frac{\frac{1}{x}\cdot x - \ln{x}\cdot 1}{x^2}dx&=\frac{\frac{1}{y}\cdot y - \ln{y}\cdot 1}{y^2}dy\\ \frac{1-\ln{x}}{x^2}dx&=\frac{1-\ln{y}}{y^2}dy\\\frac{y}{x}&=\frac{1-\ln{x}}{1-\ln{y}}\cdot\frac{y^2}{x^2}\end{align*}$$

Setting $(x, y)=(4, 2)$ to the above relation gives

$$\frac{y}{x}=\frac{1-\ln{2}}{1-\ln{4}}\times\frac{4^2}{2^2}=\boxed{\frac{4(1-\ln{2})}{1-2\ln{2}}}\approx -3.177$$