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Suppose $5x^3y -3xy^2 +y^3=6\, .$
$(1,2)$ is a point on this curve. Is the curve concave up or
concave down at $(1,2)$?

\vspace{10pt}

{\bf Explicit way to go} $y$ {\it can} be solved as a function
of $x$. Then you can differentiate the formula twice and evaluate
when $x=1$.

\vspace{10pt}

{\bf Implicit way to go} Find $\displaystyle{{dy}\over {dx}}$ implicitly and then
differentiate again to get $\displaystyle{{d^2y}\over {dx^2}}$. Evaluate
everything at $(1,2)$.


\vfill

This is what $y$ looks like as a function of $x$.

\[ \begin{split}
y =& \left( -\frac{5}{2}x^4 + 3 + x^3 + \frac{1}{18}\sqrt{1500x^9 - 675x^8 - 4860x^4 + 2916 + 1944x^3}\right)^{1/3} \\ 
&- \frac{ \frac{5}{3}x^3 - x^2}{\left( -\frac{5}{2}x^4 + 3 + x^3 + \frac{1}{18}\sqrt{1500x^9 - 675x^8 - 4860x^4 + 2916 + 1944x^3}\right)^{1/3}} + x
\end{split} \]
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