Find the extreme of $$f(x,y) = x^2+xy+y^2+y$$
I think from what I learn on the function with two variables, we need to find the second derivatives to solve this kind of question. However, after I figured out both $f_x$ and $f_y$, $$f_x=2x+y, f_y=x+2y+1$$ I am stuck on what I should do next since this function will have $f_{xx}$, $f_{xy}$, $f_{yx}$, $f_{yy}$, four second derivatives. Can someone lead me out?
You are on the rite track, but you need to do one more test before you know the extreme on this function. Clairaut's theorem:$$Df(x,y)=fxx-fyy-(fxy)^2$$
After you solve this, you may have three different result, which is $$Df(a,b)>0$$ --both fxx and fyy have same sign, so if fxx>0, you have the local max, fxx<0, you have local min; $$Df(a,b)<0$$ -- fxx and fyy have different sign, the function has no max/min, f has a saddle point; $$Df(a,b)=0$$ -- Test is inclusive.
Use this into you question and find out your answer.
If you are allowed to use graphical tools, then notice the level curve $f(x,y)=k$ is an ellipse and seems to reduce to one point when $(x,y)=(\frac 13,-\frac 23)$.
https://www.desmos.com/calculator/vymqbyrxgd
Once you have guessed that, let's try to translate the curve $f(\frac 13+X,-\frac 23+Y)=\cdots=\underbrace{X^2+XY+Y^2}_{\ge 0}-\dfrac 13$ with a minimum reached for $(X,Y)=(0,0)$.
And we have indeed a minimum $k=-\frac 13$ for the guessed point.
No maximum since trivially $f\to\infty$ at infinity.
Note that with conics you can always find the translation without hint. Just set $f(X+a,Y+b)$ and try to remove terms in $X^1$ and $Y^1$ by adjusting $a,b$.
In our case:
$f(X+a,Y+b)=\underbrace{(a^2+ab+b^2+b)}_{f(a,b)}+\underbrace{(2a+b)}_{f_x(a,b)}X+\underbrace{(a+2b+1)}_{f_y(a,b)}Y+X^2+XY+Y^2$
Thus annulating $f_x,f_y$ in that case leads to $(a,b)=(\frac 13,-\frac 23)$.
Of course here we are lucky because $X^2+XY+Y^2$ is an easy study, but in general you are left with something that is simplified (no terms in X,Y) but still need further study.
As you have obtained $f_{x}$ and $f_{y}$ one way to proceed is to look at critical points. I.e points $(a,b)$ such that $f_{x}(a,b)=f_{y}(a,b)=0$.
If you obtain such a collection of points we can then evaluate what sort of critical points these $(a,b)$ are by looking at the determinant of the Hessian matrix $$H(x,y) = \begin{bmatrix}f_{xx}(x,y) && f_{xy}(x,y)\\ f_{yx}(x,y) && f_{yy}(x,y)\end{bmatrix}.$$ Under sufficiently nice conditions on the function $f$ we have $f_{xy}=f_{yx}$ and: $$\det(H(x,y))=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2$$ Then if $\det(H(a,b))>0$: $f$ attains a $\textit{maximum}$ at $(a,b)$ if $f_{xx}(a,b)<0$ or a $\textit{minimum}$ if $f_{xx}(a,b)>0$.
If instead $\det(H(a,b))=0$ the test is inconclusive.
Lastly if $\det(H(a,b))<0$ we instead conclude that $f$ has a saddle point at $(a,b)$
I am surprised nobody mentioned the completing square approach for finding minimum (maximum is clearly infinity):
$$x^2+xy+y^2+y=\dfrac14\left(4x^2+4xy+y^2+3\left(y^2+\dfrac 43y+\dfrac 49\right)-\dfrac43\right)\\ =\dfrac14\left((2x+y)^2+3\left(y+\dfrac23\right)^2\right)-\dfrac13\ge-\dfrac13$$ Therefore, minimum value is $-\dfrac 13$ obtained at $\left(\dfrac13,-\dfrac23\right)$.
