[M@gpie]

ยุ่งยากหลายชั้นทีเดียวครับพี่ passer-by
$$ y'+ \frac{y}{x^x}(x^x \ln y)^4+y \ln x \ln y =0 $$
First, we multiply both two sides of equation by $\frac{x^x}{y}$. Then
$$ \frac{x^x}{y}\frac{dy}{dx}+ (x^x \ln y)^4+x^x \ln x \ln y =0 \; \; \; ....(*)$$
Let $u = x^x\ln y $. it is easy to compute that \[ \frac{du}{dx} - u = \frac{x^x}{y}\frac{dy}{dx} + x^x\ln x \ln y\]
(*) is changed to $$\frac{du}{dx} -u + u^4 = 0$$
This equation is the Bernoulli's equation equation, but it can be solved by seperation of variables method.