Finding mathematically the ground state density in DFT
What we have to do here is a functional derivative.
Lets consider the following method of finding functional derivartives:

$$F[\rho+\delta \rho]-F[\rho] = \int_\Omega \delta\rho \frac{\delta F[\rho]}{\delta\rho}\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{1}\label{FuncDer}$$

Lets now consider the Thomas-Fermi kinetic energy functional using the above equation.
The Thomas-Fermi functional is given as:

$$T_\mathrm{TF}[\rho] = C_\mathrm{TF}\int_\Omega\rho^{5/3}\mathrm{d}\mathbf{r}\tag{2}$$

Lets now try to evaluate $T_\mathrm{TF}[\rho+\delta\rho]$:

$$T_\mathrm{TF}[\rho+\delta\rho] = C_\mathrm{TF}\int_\Omega\left(\rho+\delta\rho\right)^{5/3}\mathrm{d}\mathbf{r}\tag{3}$$

Now lets do Taylor expansion of the integrand:

$$T_\mathrm{TF}[\rho+\delta\rho] = C_\mathrm{TF}\int_\Omega\rho^{5/3}+\frac{5}{3}\rho^{2/3}\delta\rho+\mathscr{O}(\delta \rho^2)\mathrm{d}\mathbf{r}\tag{4.1}$$ 
$$T_\mathrm{TF}[\rho+\delta\rho]= C_\mathrm{TF}\int_\Omega\rho^{5/3}\mathrm{d}\mathbf{r}+\int_\Omega C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta \rho^2)\tag{4.2}$$

Now subtracting $T_\mathrm{TF}[\rho]$:

$$T_\mathrm{TF}[\rho+\delta\rho] - T_\mathrm{TF}[\rho]=\int_\Omega C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta \rho^2)\tag{5}$$

Thus by comparing with Eq. (\ref{FuncDer}) it can be seen that:

$$\frac{\delta T_\mathrm{TF}[\rho]}{\delta\rho}=C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\tag{6}$$

Now for the second part of the kinetic energy functional in your question, i.e. the Weizsacker kinetic energy functional:

$$T_\mathrm{W}[\rho]=C\int_\Omega\frac{\nabla\rho\cdot\nabla\rho}{\rho}\mathrm{d}\mathbf{r}\tag{7}$$

Again lets evaluate $T_\mathrm{W}[\rho+\delta\rho]$:

$$T_\mathrm{W}[\rho+\delta\rho]=C\int_\Omega\frac{\nabla(\rho+\delta\rho)\cdot\nabla(\rho+\delta\rho)}{\rho+\delta\rho}\mathrm{d}\mathbf{r}\tag{8}$$

Now doing a Taylor expansion again:

$$T_\mathrm{W}[\rho+\delta\rho] = C\int_\Omega \frac{\nabla\rho\cdot\nabla\rho}{\rho} + \frac{-2\nabla^2\rho+\frac{\nabla\rho\cdot\nabla\rho}{\rho}}{\rho}\delta\rho+\mathscr{O}(\delta\rho^2)\mathrm{d}\mathbf{r}\tag{9.1}$$

$$T_\mathrm{W}[\rho+\delta\rho] = C\int_\Omega \frac{\nabla\rho\cdot\nabla\rho}{\rho}\mathrm{d}\mathbf{r} + \int_\Omega C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{9.2}$$

Now subtracting $T_\mathrm{W}[\rho]$:

$$T_\mathrm{W}[\rho+\delta\rho] - T_\mathrm{W}[\rho] = \int_\Omega C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{10}$$

Thus by comparing with Eq. (\ref{FuncDer}) it can be seen that:

$$\frac{\delta T_\mathrm{W}[\rho]}{\delta\rho} = C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\tag{11}$$

Now for the second part of your question. 
For molecular calculations involving kinetic energy functionals you have to look into orbital-free density functional theory (OF-DFT). 
The kinetic energy functionals are not used in Kohn-Sham density functional theory (KS-DFT). 
In KS-DFT the kinetic energy are calculated from the Kohn-Sham orbitals instead of the density.

Often when working with OF-DFT the equations can be solved numerically using finte element method (FEM) for a nice introduction to this take a look at Joel Davidsson master thesis [1].
Keep in mind that doing OF-DFT calculations is not a trivial task!

[1] : https://www.diva-portal.org/smash/get/diva2:864857/FULLTEXT01.pdf