Answer: When \(\frac{\mathrm{d}y}{\mathrm{d}x}=0\), the tangent is horizontal, so we first differentiate implicitly.
From \(x^3+y^3=3xy\): \(3x^2+3y^2\frac{\mathrm{d}y}{\mathrm{d}x}=3y+3x\frac{\mathrm{d}y}{\mathrm{d}x}\).
Rearranging gives \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y-x^2}{y^2-x}\), so if \(\frac{\mathrm{d}y}{\mathrm{d}x}=0\) then \(y=x^2\).
Substitute into the curve equation:
\(x^3+(x^2)^3=3x(x^2)\)
\(x^3+x^6=3x^3\)
\(x^6=2x^3\)
Since \(x\gt 0\), \(x^3=2\), so \(x=2^{1/3}\).
Then \(y=x^2=2^{2/3}\).
The turning point is \(\bigl(2^{1/3},\,2^{2/3}\bigr)\).
To determine its nature, differentiate the implicit equation again:
\(6x+3y^2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+6y\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2=3\frac{\mathrm{d}y}{\mathrm{d}x}+3x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+3\frac{\mathrm{d}y}{\mathrm{d}x}\).
At the turning point, \(\frac{\mathrm{d}y}{\mathrm{d}x}=0\), \(x=2^{1/3}\), and \(y=2^{2/3}\), so this becomes
\(6x+3y^2y''=3xy''\).
Hence
\(y''(y^2-x)=-2x\).
Using \(y^2=2x\), we get \(y^2-x=x\), so
\(y''=-2\).
Since \(y''\lt 0\), the turning point is a maximum.
Differentiate implicitly:
\(x^3+y^3=3xy\)
\(3x^2+3y^2\frac{\mathrm{d}y}{\mathrm{d}x}=3y+3x\frac{\mathrm{d}y}{\mathrm{d}x}\).
So
\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y-x^2}{y^2-x}\).
When \(\frac{\mathrm{d}y}{\mathrm{d}x}=0\), the numerator must be zero, so
\(y-x^2=0\quad\Rightarrow\quad y=x^2.\)
Substitute this into the curve equation:
\(x^3+(x^2)^3=3x(x^2)\)
\(x^3+x^6=3x^3\)
\(x^6=2x^3\)
Since \(x\gt 0\), \(x^3=2\), hence
\(x=2^{1/3}.\)
Then
\(y=x^2=2^{2/3}.\)
Therefore the turning point is \(\bigl(2^{1/3},2^{2/3}\bigr)\).
To find its nature, differentiate again:
\(6x+3y^2y''+6y(y')^2=3y'+3xy''+3y'.\)
At the turning point, \(y'=0\), so
\(6x+3y^2y''=3xy''.\)
Rearranging gives
\(6x=y''(3x-3y^2)\)
\(y''=\frac{2x}{x-y^2}.\)
Using \(y^2=2x\),
\(y''=\frac{2x}{x-2x}=-2.\)
Since \(y''\lt 0\), the turning point is a maximum.
