We will begin on the left hand side:
#y^2/x#
In order to take the derivative of this, we need to use the quotient rule:
#d/dx(u/v)=(u'v-uv')/v^2#
We have #u=y^2->u'=2ydy/dx# and #v=x->v'=1#, so:
#d/dx(y^2/x)=((2ydy/dx)(x)-(y^2)(1))/(x)^2#
#->d/dx(y^2/x)=(2xydy/dx-y^2)/x^2#
Now for the right hand side:
#x^3-3yx^2#
We can use the sum rule and multiplication of a constant rule to break this into:
#d/dx(x^3)-3d/dx(yx^2)#
The second of these will require the product rule:
#d/dx(uv)=u'v+uv'#
With #u=y->u'=dy/dx# and #v=x^2->v'=2x#. So:
#d/dx(x^3-3yx^2)=3x^2-((dy/dx)(x^2)+(y)(2x))#
#->d/dx(x^3-3yx^2)=3x^2-x^2dy/dx+2xy#
Our problem now reads:
#(2xydy/dx-y^2)/x^2=3x^2-x^2dy/dx+2xy#
We can add #x^2dy/dx# to both sides and factor out a #dy/dx# to isolate it:
#(2xydy/dx-y^2)/x^2=3x^2-x^2dy/dx+2xy#
#->(2xydy/dx)/x^2+x^2dy/dx-(y^2)/x^2=3x^2+2xy#
#->dy/dx((2xy)/x^2+x^2)=3x^2+2xy+(y^2)/x^2#
#->dy/dx=(3x^2+2xy+(y^2)/x^2)/((2xy)/x^2+x^2)#
I hope you like algebra, because this is one nasty equation that needs to be simplified:
#dy/dx=(3x^2+2xy+(y^2)/x^2)/((2xy)/x^2+x^2)#
#->dy/dx=((3x^4)/x^2+(2x^3y)/x^2+(y^2)/x^2)/((2xy)/x^2+x^4/x^2)#
#->dy/dx=((3x^4+2x^3y+y^2)/x^2)/((2xy+x^4)/x^2)#
#->dy/dx=(3x^4+2x^3y+y^2)/x^2*x^2/(2xy+x^4)#
#->dy/dx=(3x^4+2x^3y+y^2)/(2xy+x^4)#
