Curve sketching without a computer program How to sketch the curve x^6 + y^6 = (x^4)*y without using a computer program ? Could someone give me the step by step ?
 A: $\dfrac{y^6}{x^4} = y - x^2$, so there are no solutions whenever $y - x^2 < 0$. Whenever $y - x^2 = 0$, we find (by substitution) that the only solution is the origin. Notice that the curve is symmetric with respect to the y-axis, that is $(-x,y)$ is a solution if $(x,y)$ is a solution. Therefore, it is only necessary to graph the curve in the region bounded by the parabola $y = x^2$ and the y-axis. You may also want to ask if the curve is bounded in its y-component. In fact, if $y = 1$, then $x^6 - x^4 - 1 = 0$. The polynomial $f(x) = x^6 - x^4$ is pretty simple to graph with the three roots $(-1,0), (0,0),$ and $(1,0)$. By translating the graph of $x^6 - x^4$ down by one unit, to obtain the graph of $x^6 - x^4 - 1$, we see that there are two roots: one whose x-component is less than -1 and the other whose x-component is greater than 1, yet these points do not belong in the region that we described. This may be generalized for any $y \geq 1$, so the curve must be bounded above by $y = 1$.
In polar coordinate form, recall $x = r\cos(\theta)$ and $y = r\sin(\theta)$.
$x^6 + y^6 = r^6\cos^6(\theta) + r^6\sin^6(\theta) = r^6[\cos^2(\theta) + \sin^2(\theta)][\cos^4(\theta) - \cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)] = r^6[\cos^4(\theta) - \cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)]$
The equation $x^6 + y^6 = x^4y$ becomes
$r^6[\cos^4(\theta) - \cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)] = r^5\cos^4(\theta)\sin(\theta).$ We may divide appropriately since we've already discussed the points at which division would not be well-defined.
$r = \dfrac{\sin(\theta)}{1 - \tan^2(\theta) + \tan^4(\theta)}$
You may plot a few points of the form $(r,\theta)$ by way of a trigonometric table: $\big(\frac{9}{14}, \frac{\pi}{6}\big), \big(\frac{1}{\sqrt{2}},\frac{\pi}{4}\big), \big(\frac{\sqrt{3}}{14}, \frac{\pi}{3}\big)$.
A: If $x=0$ or $y=0$ the given equation enforces $x=y=0$. Therefore assume $xy\ne0$. Then the equation is equivalent with
$${x^2\over y}+{y^5\over x^4}=1\ .$$
Put
$${x^2\over y}=:u,\quad{y^5\over x^4}=:v\ .$$
Then on the one hand $u+v=1$, and on the other hand
$$x^6=u^5 v,\quad y^3=u^2 v\ .$$
This leads to the following "formal" parametric representation of the solution set:
$$u\mapsto(x,y):=\left(\root 6\of{u^5(1-u)},\> \root3\of{u^2(1-u)}\>\right)\qquad(-\infty<u<\infty)\ ,\tag{1}$$ 
whereby you will obtain $\geq0$ admissible values, depending on the choice of $u$ and the sign of the sixth root. At any rate, the representation $(1)$ allows to produce  arbitrarily many points of the solution set with the help of a pocket calculator.
