Is there any point #(x, y)# on the curve #y=x^(x(1+1/y)), x > 0,# at which the tangent is parallel to the x-axis?

First, let's consider the conditions of the tangent if it is parallel to the #x#-axis. Since the #x#-axis is horizontal, any line parallel to it must also be horizontal; so it follows that the tangent line is horizontal. And, of course, horizontal tangents occur when the derivative equals #0#.

Therefore, we must first start by finding the derivative of this monstrous equation, which can be accomplished through implicit differentiation:
#y=x^(x+x/y)#
#->lny=(x+x/y)lnx#

Using the sum rule, chain rule, product rule, quotient rule, and algebra, we have:
#d/dx(lny)=d/dx((x+x/y)lnx)#
#->dy/dx*1/y=(x+x/y)'(lnx)+(x+x/y)(lnx)'#
#->dy/dx*1/y=(x+x/y)'(lnx)+(x+x/y)(lnx)'#
#->dy/dx*1/y=(1+(x'y-xdy/dx)/y^2)(lnx)+(x+x/y)(1/x)#
#->dy/dx*1/y=lnx+lnx((y-xdy/dx)/y^2)+1+1/y#
#->dy/dx*1/y=lnx+lnx(1/y-(xdy/dx)/y^2)+1+1/y#
#->dy/dx*1/y=lnx+(lnx)/y-(xlnxdy/dx)/y^2+1+1/y#
#->dy/dx*1/y+(xlnxdy/dx)/y^2=lnx+(lnx)/y+1+1/y#
#->dy/dx(1/y+(xlnx)/y^2)=lnx+(lnx)/y+1+1/y#
#->dy/dx((y+xlnx)/y^2)=lnx+(lnx)/y+1+1/y#
#->dy/dx((y+xlnx)/y^2)=(ylnx+lnx+1+y)/y#
#->dy/dx=((ylnx+lnx+1+y)/y)/((y+xlnx)/y^2)#
#->dy/dx=(y(ylnx+lnx+1+y))/(y+xlnx)#

Wow...that was intense. Now we set the derivative equal to #0# and see what happens.
#0=(y(ylnx+lnx+1+y))/(y+xlnx)#
#0=ylnx+lnx+1+y#
#-ylnx-y=lnx+1#
#-y(lnx+1)=lnx+1#
#y(lnx+1)=-(lnx+1)#
#y=(-(lnx+1))/(lnx+1)#
#y=-1#

Interesting. Now let's plug in #y=-1# and see what we get for #x#:
#y=x^(x(1+1/y))#
#-1=x^(x(1+1/-1))#
#-1=x^(x(1-1))#
#-1=x^0#
#-1=1#

Since this is a contradiction, we conclude that there are no points meeting this condition.

#y =x^(x (1 + 1/y)) equiv y^{y/(y+1)} = x^x#. Now calling #f(x,y) = x^x-y^{y/(y+1)} = u(x) + v(y) = 0# we have

#df = f_x dx + f_y dy = (partial u)/(partial x) dx + (partial v)/(partial y) dy = 0# then

#dy/dx = -( (partial u)/(partial x))/((partial v)/(partial y)) = (x^x (1 + Log_e(x))(1 + y)^2)/(y^(y/(1 + y)) (1 + y + Log_e(y))) = ((1 + Log_e(x))(1 + y)^2)/ (1 + y + Log_e(y))#

We see that #dy/(dx)=0 -> {y_0 = -1, x_0 = e^{-1}}# but those values must verify:

#f(x,y_0) = 0# and
#f(x_0,y) = 0#

In the first case, #y_0 = 1# we have

#x^x = -1# which is not attainable in the real domain.

In the second case, #x_0 = e^{-1}# we have

#y^{y/(y+1)} = e^{-1}# or
#y/(y+1)log_e y = -1#

but

#y/(y+1)log_e y > -1# so no real solution also.

Concluding, there is not such a tangent.

I had proposed this question to get this value precisely. Thanks to

Dr, Cawas for a decisive answer that approves the revelation that

the double precision y' remains 0 around this interval. y is

continuous and differentiable at x = 1/e. As both the 17-sd double

precision y and y' are 0, in this interval around x = 1/e, it was a

conjecture that x-axis touches the graph in between. And now, it is

proved. I think that the touch is transcendental. .