y=x^(e^c*x)
:. ln y=ln{x^(e^c*x)}=(e^c*x)(lnx).
:. d/dx{lny}=d/dx{(e^c*x)(lnx)}.
Applying the Chain Rule for the L.H.S., and, the
Product Rule for the R.H.S., we get,
{d/dy(lny)}{dy/dx}=(xe^c)d/dx(lnx)+(lnx)d/dx((xe^c)).
:. 1/ydy/dx=(xe^c)(1/x)+(lnx)(e^cd/dx(x)).
=e^c+(e^c)(lnx)(1)=e^c(1+lnx)
rArr dy/dx=e^cy(1+lnx)..............(d_1).
Next, (d^2y)/dx^2=d/dx{dy/dx},
=d/dx{e^cy(1+lnx)},
=e^cd/dx{y(1+lnx)},
=e^c[yd/dx(1+lnx)+(1+lnx)d/dx(y)],
=e^c{y*1/x+(1+lnx)e^cy(1+lnx)}....[because, (d_1)],
=y/x*e^c{1+xe^c(1+lnx)^2}.
Since, y=x^(e^c*x), we have,
(d^2y)/dx^2=e^cx^(e^c*x-1){1+xe^c(1+lnx)^2}.
Enjoy Maths.!