No. Consider the case of a surjective bounded linear operator $T:X\to Y$ which is not a (top-linear) left inverse (that is, $\operatorname{ker}(T)$ does not split in $X$). However by classical selection theorems a surjective bounded linear operator $T$ has a continuous right inverse  $g$, even $1$-homogeneous; but it can be  differentiable   at no point, otherwise differentiating you would get a bounded linear right inverse to $T$.