The domain of $y$ is $x \in [0, 13]$. We have
$$\begin{aligned} y &= \sqrt{x+27} + \sqrt{13-x} + \sqrt{x} \\ &= \sqrt{x+27} + \sqrt{13 + 2\sqrt{x(13-x)}} \\ &\geq \sqrt{27} + \sqrt{13} = 3\sqrt{3} + \sqrt{13}. \end{aligned}$$
The equality holds when $x = 0$. Therefore, the minimum of $y$ is $3\sqrt{3} + \sqrt{13}$.

On the other hand, by the Cauchy inequality we have
$$\begin{aligned} y^2 &= (\sqrt{x} + \sqrt{x+27} + \sqrt{13-x})^2 \\ &\leq \left(\frac{1}{2} + 1 + \frac{1}{3}\right) \left[2x + (x+27) + 3(13-x)\right] \\ &= 121. \end{aligned}$$
The equality holds when $4x = 9(13-x) = x + 27$. It is so for $x = 9$. Therefore, the maximum of $y$ is $11$.