#2y^2=x^3-5#

#y^2=1/2x^3-5/2#

#=>y=+-(1/2x^3-5/2)^(1/2)#

Let #u= 1/2x^3-5/2" " =>" " (du)/(dx)=3/2x^2#

Consider:

#y=+-u^(1/2)" " =>" " dy/(du)=1/2u^(-1/2)#

But #dy/dx=(du)/dxxxdy/(du)#

#=>dy/dx=1/2u^(-1/2)xx3/2x^2#

#=>dy/dx=1/2(1/2x^3-5/2)^(-1/2)xx3/2x^2#

#=>dy/dx=3/4x^3(1/2x^3-5/2)^(-1/2)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Second differential

Let #u=3/4x^3" " =>" " (du)/dx=9/4x^2#

Let #v=(1/2x^3-5/2)^(-1/2)#

#=>" "(dv)/dx=-1/2(1/2x^3-5/2)^(-3/2)(3/2x^2)#

#=>" "(dv)/dx=-3/4x^2(1/2x^3-5/2)^(-3/2)#

So for the second differential we have:

#(d^2y)/(dx^2) -> v (du)/dx+u(dv)/(dx)#

#color(white)(.)#

#(d^2y)/(dx^2) ->[(1/2x^3-5/2)^(-1/2)xx9/4x^2] + [3/4x^3xx(-3/4x^2(1/2x^3-5/2)^(-3/2)]#

#color(white)(.)#

#(d^2y)/(dx^2) ->(9x^2)/(4xxsqrt(1/2x^3-5/2))- (9x^5)/(16xx(sqrt(1/2x^3-5/2))^3)#

I will let you take it from this point