Smooth function vanishing on axis

Since $f$ is smooth and vanishes on the $x-$axis, we have $$f(x,y) = \int_0^y f_y(x,s) ds.\tag{1}$$ Because $f$ vanishes on the $y-$axis, we know that $f_y=0$ on the $y-$axis; so if we restrict to the closed ball $\overline{B(0,1)}$ so that $|f_{xy}|$ is bounded by some constant $k$ then we have $$|f_y(x,y)|=\left|\int_0^xf_{xy}(s,y)ds\right|\le \int_0^xk\,ds =k|x|.$$ Thus, applying the exact same kind of estimate to $(1)$ we conclude $$|f(x,y)|\le k|x||y|$$ as desired.