Figure

$f(x,y)=3x^{2}-xy+y^{2}$ is a homogeneous function of degree 2, since $\begin{eqnarray*} f(tx,ty)& =&3(tx)^{2}-( tx) ( ty) +(ty)^{2}=t^{2}3x^{2}-t^{2}xy+t^{2}y^{2} \\ & =&t^{2}(3x^{2}-xy+y^{2})=t^{2}f(x,y)\quad \hbox{for all }t>0 \end{eqnarray*}$
$f(x,y)=\sqrt{x+4y}$ is a homogeneous function of degree $\dfrac{1}{2}$ , since $\begin{eqnarray*} f(tx,ty)&=&\sqrt{tx+4( ty) }=\sqrt{t(x+4y)}=\sqrt{t}\sqrt{x+4y}\\ &=&t^{1/2}f(x,y)\quad \hbox{for all }t>0 \end{eqnarray*}$
$f(x,y)=\dfrac{x}{\sqrt{x^{2}-y^{2}}}$ is a homogeneous function of degree $0$ , since $\begin{eqnarray*} f(tx,ty)&=&\frac{tx}{\sqrt{(tx)^{2}-(ty)^{2}}}=\frac{tx}{\sqrt{t^{2}(x^{2}-y^{2})}}=\frac{tx}{t\sqrt{x^{2}-y^{2}}}\\ &=&t^{0}f(x,y)\quad \hbox{for all }t>0 \end{eqnarray*}$
$f(x,y)=x-y^{2}$ is not a homogeneous function, since $f(tx,ty)=tx-(ty)^{2}=t(x-ty^{2})\neq t^{k}(x-y^{2})$ for all $t>0$ and some $k$