Simplify the expression noting that based on properties of logarithms:

#f(x) = e^x ln(x^2) = 2e^xlnabsx#

the critical points are by definition the points where #f'(x) = 0# so evaluate the derivative of the function:

#(df)/dx = d/dx (2e^xlnabsx) =2 ((d/dx e^x)lnabsx +e^x (d/dx lnabsx))#

#(df)/dx =2 (e^xlnabsx +e^x/x)#

#(df)/dx =2 e^x(lnabsx +1/x)#

To identify the critical points then we have to solve the equation:

#2 e^x(lnabsx +1/x) = 0#

and as #2e^x != 0#:

#lnabsx=-1/x#

This is a logarithmic equation that can be solved only approximately:

note that for #x>0# we have #lnx = -1/x < 0# so the solution should be in the interval #(0,1)#.

However given:

#lnx+1/x = 0#

as #x !=0# this is equivalent to:

#xlnx + 1 = 0#

Now consider the function #g(x) = xlnx+1#:

#lim_(x->0) (xlnx+1) = 1+ lim_(x->0) xlnx#

#lim_(x->0) (xlnx+1) = 1+ lim_(x->0) lnx/(1/x)#

Apply l'Hospital's:

#lim_(x->0) (xlnx+1) = 1+ lim_(x->0) (d/dx lnx)/(d/dx 1/x)#

#lim_(x->0) (xlnx+1) = 1+ lim_(x->0) (1/x)/(-1/x^2)= 1+ lim_(x->0) (-x) = 1#

while:

#lim_(x->1) (xlnx+1) = 1#

Consider now:

#d/dx (xlnx+1) = 1+lnx#

so the only critical point for this function is in #x=1/e# where its value is:

#g(1/e) = 1/eln(1/e)+1 = 1-1/e > 0#

and as #g''(1/e) = e >0# this is a local minimum. As the minimum value of the function is positive, this means that it is never zero. So actually we have no critical points for #x > 0#

On the other hand for #x < 0# the equation becomes:

#xln(-x) +1 =0#

and substituting #t=-x# we can see that:

#lim_(x->0^-) (xln(-x)+1) = 1- lim_(t->0^+) tlnt =1#

while:

#lim_(x->-oo) (xln(-x)+1) = -oo#

thus the function assumes positive and negative values and as it is continuous in the interval #(-oo,0)# it must vanish at least in one point.

graph{e^x ln(x^2) [-10, 10, -5, 5]}