How do you differentiate #-1=y^3/(1+x^2y)#?

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Answer 1 · 2016-12-14T12:04:48

#-(2xy)/(3y^2+x^2), y < 0 and x in (-oo, oo)#.

#y<0#, to make #x^2 >0#.

Rearranging,

#y^3+yx^2+1=0#. So,

#3y^2y'+x^2y'+2xy=0#. And so,

#y'=-(2xy)/(3y^2+x^2), y < 0 and x in (-oo, oo)#

#=0, # at #(0, -1)#

graph{y(x^2+y^2)+1=0 [-6.37, 6.365, -3.185, 3.185]}