Let $x=r\cos \theta, \quad y=r\sin \theta$
Therefore $$|\frac{xy}{\sqrt{x^2+y^2}}|=r |\cos \theta \sin \theta|\le r=\sqrt{x^2+y^2}\lt \epsilon$$if $x^2\lt\frac{\epsilon^2}{2},\quad$and$\quad y^2\lt \frac{\epsilon^2}{2}$
or if $|x|\lt\frac{\epsilon}{\sqrt2},\quad$and$\quad |y|\lt \frac{\epsilon}{\sqrt2}$
Thus $$|\frac{xy}{\sqrt{x^2+y^2}}-0|\lt \epsilon,\qquad \text{where}\quad |x|\lt\frac{\epsilon}{\sqrt2},\quad and \quad |y|\lt \frac{\epsilon}{\sqrt2}$$
$$\implies \lim_{(x,y)\to (0,0)}\frac{xy}{\sqrt{x^2+y^2}}=0$$
Hence $$\lim_{(x,y)\to (0,0)}f(x,y)=f(0,0)$$and therefore $f(x,y)$ is continuous at $(0,0)$
Again $$f_x(0,0)=\lim_{h \to 0}\frac{f(h,0)-f(0,0)}{h}=0$$and $$f_y(0,0)=\lim_{k \to 0}\frac{f(0,k)-f(0,0)}{k}=0$$
Thus the function $f(x,y)$ possesses partial derivatives at $(0,0)$.
If the given function is differentiable at $(0,0)$, then by definition $$df=f(h,k)-f(0,0)=Ah+Bk+h\phi +k\psi\qquad . . . . . (1)$$ where $\quad A=f_x(0,0)=0;\quad B=f_y(0,0)=0, \quad $and $~\phi,~\psi~$ tends to zero as $\quad (h,k)\to (0,0)$.
So from $(1)$ we have $$\frac{hk}{\sqrt{h^2+k^2}}=h\phi +k\psi$$
Putting $\quad k=mh \quad $and letting $\quad h\to 0,\quad$ we have
$$\frac{m}{\sqrt{1+m^2}}=\lim_{h \to 0}(\phi +m\psi)=0$$which is impossible for arbitraty $~m$.
Hence the function is not differentiable at $(0,0)$.
Best Answer
As noted by @geetha290krm, you can observe that $f(3)>5$ and $f(0)<5$. But notice that you don't have to explicitly find those specific points. Since you know that $\lim\limits_{x\to+\infty}f(x)=+\infty$, which means that $\forall M\in\mathbb{R}\ \exists k\in\mathbb{R}\ \forall x\geq k:f(x) > M$, you have that $\exists k\in\mathbb{R}\ \forall x\geq k:f(x)>5$ by taking $M=5$. Similarly, since you know $\lim\limits_{x\to-\infty}f(x)=-\infty$, you can find $\exists k'\in\mathbb{R}\ \forall x\leq k':f(x)<5$. Hence, since $f(k')<5$ and $f(k)>5$, you know by continuity that $\exists x\in(k',k):f(x)=5$. This is the formal reason why you can indeed apply the IVT on the limiting case (it is actually rigorous).