$g \in C^{1}(\Bbb{R}^{n})$, $g$ and $Dg$ are bounded in $\Bbb{R}^{n}$.
$$u(x,t) = \frac{1}{(4\pi t)^{\frac{n}{2}}} \int_{\Bbb{R}^n} e^{-\frac{|x-y|^2}{4t}} g(y) dy$$ $(x,t) \in \Bbb{R}^n \times (0,\infty)$
I am trying to show that $$\sup_{\Bbb{R}^{n} \times (0,\infty)} |u_{x_{i}}| \leq \sup_{\Bbb{R}^n} |g_{x_{i}}|$$ for $i=1,\ldots n$
I was thinking to take absolute value on both sides of the expression of $u(x,t)$ and trying some tricky integration inequality might help?
Observe that your $u$ is the solution of the IVP for the heat equation $$ u_t=\Delta u $$ with initial data $u(x,0)=g(x)$. Then $|u(x,t)|\le Sup\,|g|$ (comparison principle). You can check this by taking the integral of the absolute value and observing that $$ \int\limits_{\mathbb{R}^n}e^{-\frac{|x-y|^2}{4t}}\,dy=(4\pi t)^{n/2}. $$ for every $x$ and $t>0$.
Last, you observe that $u_{x_i}(x,t)$ is the solution of the initial value problem for the same heat equation with initial datum $u_{x_i}(x,0)=g_{x_i}(x)$ (just take the derivative with respect to $x_i$ in the equation and exchange the order of derivatives). Then, the same comparison principle applies and gives the result you need.
Obs. The solution is unique for bounded data, therefore it is the one given by the convolution with the heat kernel.