A counterexample: $f(x,y):=x^2y^2$ if $xy\ne0$ and $f(x,y):=|x|+|y|$ otherwise.
The OP has changed the question to additionally require that $f$ be smooth. Here is a smooth counterexample: $f(x,y):=x^2y^2+(x^2+y^2)e^{-x^2y^2(x^2+y^2)}$.
Indeed, take any $(a,b)\in S^1$. If $ab\ne0$, then $f(ta,tb)\ge t^2(ab)^2\to\infty$ as $t\to\infty$. If $a=0$, then $b\ne0$ and $f(ta,tb)=t^2b^2\to\infty$ as $t\to\infty$. The case $b=0$ is similar. However, if $0<x\to\infty$ and $y=x^{-3/2}$, then $$f(x,y)=\frac1x+\Big(x^2+\frac1{x^3}\Big) \exp\Big(-x-\frac1{x^4}\Big)\to0.$$