Examine the function \(f(x,y)=y^2+4xy+3x^2+x^3\) for extreme values.

\(f_x=4y+6x+3x^2,\quad f_y=2y+4x\).

Stationary points: \(f_y=0\Rightarrow y=-2x\). Then \(f_x=-8x+6x+3x^2=3x^2-2x=0\Rightarrow x=0\) or \(x=\frac23\).

Points: \((0,0)\) and \(\left(\frac23,-\frac43\right)\).

\(r=f_{xx}=6+6x,\quad s=f_{xy}=4,\quad t=f_{yy}=2\).

At (0, 0): \(r=6,\ rt-s^2=12-16=-4<0\) → saddle point (no extreme value).

At \(\left(\frac23,-\frac43\right)\): \(r=10,\ rt-s^2=20-16=4>0,\ r>0\) → minimum.

\(f_{min}=\frac{16}9+4\cdot\frac23\cdot\left(-\frac43\right)+3\cdot\frac49+\frac8{27}=\frac{16-32+12}{9}+\frac8{27}=-\frac{12}{27}+\frac8{27}\)

Minimum value \(=-\frac4{27}\) at \(\left(\frac23,-\frac43\right)\); no maximum.