Assume that $X$ and $Y$ are Banach spaces and $T:X\to Y$ is a bounded surjective linear map. Is there a Gateaux differentiable function $g:Y\to X$ such that $T\circ g=Id_{Y}$?

No. Consider the case of a surjective bounded linear operator $T:X\to Y$ which is not a (toplinear) 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 Gdifferentiable at no point, otherwise differentiating you would get a bounded linear right inverse to $T$. Rmk. I refer to the standard definition of the Gâteaux differential of $f:X \to Y$ at $x$, that is, a bounded linear operator $L$ such that for all $v\in X$ there holds $\frac{d}{dt}f(x+tv)\Big_{t=0}=Lv$. In particular the chain rule holds. (Warning: some adopt a weaker definition, where $L$ is not even linear.) 

