Is the second partial derivative test inconclusive? Is the second partial derivative test, $f_{xx}=0$, inconclusive? If yes, I am confused as to why my textbook says this function:
$$f(x,y)=x^2+y^2+x^2y+4$$
Has a minimum point at $(0,0)$. When I use the test, It ends up as $D=4>0$ and $f_{xx}=0$, so how I am supposed to get to the conclusion it is a minimum point?
 A: Apparently this was your first question. So well done! MSE can be intimidating at first.
Well $f_{xx}=2+2y$, so $f_{xx}(0,0)=2>0$.
There is quite a reasonable Wikipedia article on the test.
You probably made an error calculating $f_{xx}$. I am afraid you were not helped by the initial edit carried out for you on the question. But then your original question was a little confusing. Good formatting really helps!
Incidentally, this function also illustrates the dangers of math software. Here is a plot from Mathematica:

At a quick glance it looks as though it might be a saddle point. But if you look closely, maybe not. But if you do a 2D plot along any line through the origin, it has a minimum at the origin (not that that is conclusive).
If you need further help please comment below this answer.
A: When unsure, convert the function to polar coordinates around the point $(x_0, y_0)$ being examined, using
$$\begin{cases}
x = x_0 + r \cos\theta \\
y = y_0 + r \sin\theta \\
\end{cases}$$
In this case, around $(0, 0)$,
$$\begin{array}{rcl}
f(r, \theta) & = & (r \cos \theta)^2 + (r \sin \theta)^2 + (r \cos \theta)^2 (r \sin \theta) + 4 \\
~ & = & r^2 + r^3 (\cos\theta)^2 (\sin\theta) + 4
\end{array}$$
Next, find the minimum and maximum contribution from the angular component $\theta$.  In this case,
$$-\frac{2}{3 \sqrt{3}} \le (\cos\theta)^2 (\sin\theta) \le \frac{2}{3 \sqrt{3}}$$
i.e. $-0.3849 \le (\cos\theta)^2(\sin\theta) \le +0.3849$.
(These correspond to the directions $\theta$ where the surface changes most rapidly.)
Thus, the entire surface (around point $(0,0)$) is bounded between $f_{min}$ and $f_{max}$,
$$\begin{aligned}
f_{min}(r) = r^2 - \frac{2}{3 \sqrt{3}} r^3 + 4 \\
f_{max}(r) = r^2 + \frac{2}{3 \sqrt{3}} r^3 + 4 \\
\end{aligned}$$
If you draw $f_{min}(r)$ and $f_{max}(r)$ for $r \ge 0$, you'll see they both curve upwards from 4!
This means that there indeed is a small dip or valley at $(0,0)$.
You can clearly see this if you plot the original function $f(x,y) = x^2 + y^2 + x^2 y + 4$ close to origin, say $(-1,-1)-(+1,+1)$.
In fact, plotting region $(-1,-1)-(+1,+1)$, and comparing it to first $(-2,-2)-(+2,+2)$ and then to $(-10,-10)-(+10,+10)$ is an excellent example of how important it is to look at the problem at different scales to see which one makes most intuitive sense.  You can use Gnuplot for interactive examination,
set isosamples 20 ; set hidden3d
set xrange [-1:+1] ; set yrange [-1:+1] ; splot x*x+y*y+x*x*y+4 notitle w surface

or e.g. Wolfram Alpha online.
