【正确答案】[解] 由
df(x，y)=ex²yd(x²+y)+(x²+y)e^x2yd(x²y)
=e^x2y(2x²dx+dy)+(x²+y)e^x2y(2xydx+x²dy)
=e^x2y[2x+2(x²+y)xy]dx+e^x2y[1+(x²+y)x²]dy，可得
f'_x(x，y)=e^x2y[2x+2xy(x²+y)]，
f'_y(x，y)=e^x2y[1+x²(x²+y)]，所以
f'_x(x，2x)=2(x+4x³+2x⁴)e^2x3，
f'_y(x，2x)=(1+2x³+x⁴)e^2x3，也可用链式法则计算