fx=2x(2+y²)fy=2yx²+lny+y* 1/y =2yx²+lny+1令fx0,fy=0，解得：x=0,y=1/e所以f(x)在点(0,1/e)取得极值。极值=0+(1/e)*[ln(1/e)]=-1/e
