Range of $f(x,y)=\frac{4x^2+(y+2)^2}{x^2+y^2+1}$

See fig. below.

$5$ is the maximum. More precisely :

$$f(x,y):=\dfrac{4x^2+(y-2)^2}{x^2 + y^2 + 1}\leq 5 \tag{1}$$

with case of equality iff

$$(x,y)=(0,1/2)\tag{2}$$

Here is why : we form the "gap" between the LHS and RHS of (1) :

$$5-\dfrac{4x^2+(y-2)^2}{x^2 + y^2 + 1}=\dfrac{x^2+(2y - 1)^2}{x^2 + y^2 + 1}$$

This gap is always >0 with the exceptional case given by (2).

Besides, $0$ is clearly the minimum, realized for $(x,y)=(0,2)$.

Now the last question : is the range of values of function $f$ the whole interval $[0,5]$ ? Answer : yes by continuity.

Fig. 1 : Surface $z=f(x,y)$ with its unique minimum and maximum.

For $$f(x,y)=\frac{4x^2+(y+2)^2}{x^2+y^2+1}$$ partial derivatives are not so terrifying: \begin{align} D_1f(x,y)&=\frac{8x(x^2+y^2+1)-2x(4x^2+(y+2)^2)}{(x^2+y^2+1)^2}\\[4px] &=\frac{2xy(3y-4)}{(x^2+y^2+1)^2} \\[8px] D_2f(x,y)&=\frac{2(y+2)(x^2+y^2+1)-2y(4x^2+(y+2)^2)}{(x^2+y^2+1)^2}\\[4px] &=\frac{2(-3x^2y+2x^2-2y^2-3y+2)}{(x^2+y^2+1)^2} \end{align} The first derivative vanishes for $x=0$, $y=0$ or $y=4/3$.

We have (the denominator is irrelevant as it doesn't vanish) \begin{align} D_2f(0,y)&=\frac{-2(2y^2+3y-2)}{\dots} \\[4px] D_2f(x,0)&=\frac{4(x^2+1)}{\dots} \\[4px] D_2f(x,4/3)&=\frac{-4(x^2+19/3)}{\dots} \end{align} Thus we only get critical points for $x=0$ and $2y^2+3y-2=0$, that is, $y=-2$ or $y=1/2$.

Since $f(0,-2)=0$ and $f(0,1/2)=5$, the minimum is $0$ and the maximum is $5$.

Well, how do we know that $0$ is the minimum? Because obviously the function only takes on nonnegative values. Why $(0,5)$ is a point of maximum and not a saddle point? Because $f(x,y)$ is bounded: $$ f(x,y)\le\frac{4x^2+4(y+2)^2}{x^2+y^2+1}=4+\frac{16y+12}{x^2+y^2+1}\le16+\frac{16y}{x^2+y^2+1} $$ Now note that $|y|\le x^2+y^2+1$, because $|y|^2-|y|+1\ge0$. Thus $$ |f(x,y)|\le 32 $$ Hence the function must have an absolute maximum which has to be at a critical point. Checking with the Hessian would be indeed a nuisance.