Find points on a curve when slope is 0

MrJamesta
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Homework Statement
Find the points on the curve xy^2+x^2y=16 when the tangent line is horizontal

The attempt at a solution
I found the derivative of the curve
-(y(2x+y))/(x(x+2y))
then I found what values of y make the derivative equal 0
y=0,-2x
Then I went to plug into the original curve to find x and that is where I got lost. I do not know how to isolate x in this case.
 
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MrJamesta said:
Homework Statement
Find the points on the curve xy^2+x^2y=16 when the tangent line is horizontal

The attempt at a solution
I found the derivative of the curve
-(y(2x+y))/(x(x+2y))
then I found what values of y make the derivative equal 0
y=0,-2x
Then I went to plug into the original curve to find x and that is where I got lost. I do not know how to isolate x in this case.

The first thing you should notice is that ##y=0## doesn't give a point on the graph. But what happens if you put ##y=-2x## into the original equation?
 
Thanks! Got it now.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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