This is a classic example of implicit differentiation.
This is almost the same as taking the derivate with only one term #(frac(d)(dx))#, but everytime there is a #y#, we write #frac(dy)(dx)#, as we will be taking the derivative of "#y# with respect to #x#." In other words, if there is a #y# and an #x# together, we must apply the product rule (#y# will follow the same derivation as #x#).
The derivative of #y^3*e^x# is #frac(d)(dx)(y^3*e^x)=y^3*e^x+3y^2e^xfrac(dy)(dx)#, by the product rule.
The derivative of #-x^2# is #-2x#.
The derivative of #5*e^y# is #5*e^y(frac(dy)(dx))#, as the #y# is not with an #x# term, so we cannot treat it as a constant.
And of course, the derivative of #1# is #0#.
Now we can rearrange our equation so that the #frac(dy)(dx)# is isolated.
#y^3*e^x+3y^2e^xfrac(dy)(dx)-2x+5*e^y(frac(dy)(dx))=0#
#3y^2e^x(frac(dy)(dx))+5*e^y(frac(dy)(dx))=2x-y^3*e^x#
#frac(dy)(dx)=frac(2x-y^3*e^x)(3y^2e^x+5*e^y)#.
Hopefully I didn't do anything silly, and that this helps. Good luck :).
