
Differentiation question
The equation of a curve is xy(x+y)=2(a)^3 where a is a nonzero constant. show that there is only one point on the curve at which the tangent is parallel to the xaxis, and find the coordinates of this point.
can some1 please help me with this question please.
thanks in advance

$\displaystyle \begin{array}{ccl}xy(x+y) & = & 2a^3 \\ x^2y + xy^2 & = & 2a^{3} \\ {\color{red}(x^2y)'} + {(\color{blue} xy^2)' } & = & {\color{magenta}(2a^3)'} \\ {\color{red}\left[(x^2)'y + x^2(y)'\right]} + {\color{blue} \left[ (x)'y^2 + x(y^2)'\right]} & = & {\color{magenta}0} \end{array}$
So far, I've multiplied in the xy into the brackets. Then, I differentiate both sides. Since y is a function of x, we have to use product rule on each product on the left hand side. On the right hand side, since a is just a constant, its derivative is simply 0.
See where you can go from this and come back with any more questions.