$$\displaystyle\lim_{h\to0}\frac{e^{(x+h)\ln(x+h)}-e^{x\ln(x)}}{h} = \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{(x+h)\ln(x+h)- x \ln(x)}-1}{h}$$

$$ = \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x\ln(\frac{x+h}{x})+h \ln(x+h)}-1}{h} \quad = \quad \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x\ln(\frac{x+h}{x})} e^{h \ln(x+h)}-1}{h} $$

$$\displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x\ln(1+\frac{h}{x})}e^{h \ln(x+h)}-1}{h} =^{taylor} \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x \frac{h}{x}}(x+h)^h-1}{h} $$

$$x^x \displaystyle\lim_{h\to0}\frac{e^h (x+h)^h-1}{h} $$

That's how far I'm able to get without using chain rules or similar. Can anyone help me conclude?