A maxima occurs when the derivative of the function equals zero and second derivative is negative. Hence we first find out the derivative of #x/(1+xtanx)# for which we use quotient rule.

According to quotient rule if #Y=(f(x))/(g(x))#

then #(dy)/(dx) = (g(x)(df(x))/(dx)-f(x)(dg(x))/(dx))/((g(x))^2)#

Here #f(x)=x# and #(df)/(dx)=1# and

#g(x)=1+xtanx# and #(dg)/(dx)=tanx+xsec^2x#

Hence #(dy)/(dx) = (1+xtanx-x(tanx+xsec^2x))/(1+xtanx)^2#

= #(1-x^2sec^2x)/(1+xtanx)^2#

and #(dy)/(dx)=0# when #1-x^2sec^2x=0# or

#x^2=cos^2x# or #x=+-cosx#

We now workout second derivative

#(d^2y)/(dx^2)=((1+xtanx)^2(-2xsec^2x-2x^2sec^2xtanx)-2(1-x^2sec^2x)(1+xtanx)(tanx+xsec^2x))/(1+xtanx)^4#

amd when #x=+-cosx#

#(d^2y)/(dx^2)=(-2(1+-sinx)^2(x^3+tanx))/(1+xtanx)^4#

As #(d^2y)/(dx^2)<=0# when #x=cosx# not at #x=-cosx#

We have a local maxima at #x=cosx# and graph gives the solution as #x=0.739# as appears from graph below. See the point of intersection of #y=x# and #y=cosx# is at about #x=0.739#.

graph{(y-x)(y-cosx)=0 [-0.667, 1.833, -0.13, 1.12]}

Graph of #x/(1+xtanx)# as it appears around local maxima is given below.

graph{x/(1+xtanx) [-0.0256, 1.2155, -0.0604, 0.5603]}