Answer
$$\frac{\partial}{\partial x}\phi(x,y,z,t)=\frac{\alpha}{\gamma z+\delta t^2}.$$
$$\frac{\partial}{\partial y}\phi(x,y,z,t)=\frac{2\beta y}{\gamma z+\delta t^2}.$$
$$\frac{\partial}{\partial z}\phi(x,y,z,t)=-\frac{\gamma(\alpha x+\beta y^2)}{(\gamma z+\delta t^2)^2}.$$
$$\frac{\partial}{\partial t}\phi(x,y,z,t)=-\frac{2\delta t(\alpha x+\beta y^2)}{(\gamma z+\delta t^2)^2}.$$
Work Step by Step
Regard $y,z$ and $t$ as constant and differentiate with respect to $x$:
$$\frac{\partial}{\partial x}\phi(x,y,z,t)=\frac{\partial}{\partial x}\left(\frac{\alpha x+\beta y^2}{\gamma z +\delta t^2}\right)=\frac{1}{\gamma z+\delta t^2}\frac{\partial}{\partial x}(\alpha x+\beta y^2)=\frac{\alpha}{\gamma z+\delta t^2}.$$
Regartd $x, z$ and $t$ as constant and differentiate with respect to $y$:
$$\frac{\partial}{\partial y}\phi(x,y,z,t)=\frac{\partial}{\partial y}\left(\frac{\alpha x+\beta y^2}{\gamma z +\delta t^2}\right)=\frac{1}{\gamma z+\delta t^2}\frac{\partial}{\partial y}(\alpha x+\beta y^2)=\frac{1}{\gamma z+\delta t^2}\beta\cdot2y=\frac{2\beta y}{\gamma z+\delta t^2}.$$
Regard $x, y$ and $t$ as constant and differentiate with respect to $z$:
$$\frac{\partial}{\partial z}\phi(x,y,z,t)=\frac{\partial}{\partial z}\left(\frac{\alpha x+\beta y^2}{\gamma z +\delta t^2}\right)=(\alpha x+\beta y^2)\frac{\partial}{\partial z}\left(\frac{1}{\gamma z+\delta t^2}\right)=(\alpha x+\beta y^2)\frac{-1}{(\gamma z+\delta t^2)^2}\frac{\partial}{\partial z}(\gamma z+\delta t^2)=-\frac{\gamma(\alpha x+\beta y^2)}{(\gamma z+\delta t^2)^2}.$$
Regard $x, y$ and $z$ as constant and differentiate with respect to $t$:
$$\frac{\partial}{\partial t}\phi(x,y,z,t)=\frac{\partial}{\partial t}\left(\frac{\alpha x+\beta y^2}{\gamma z +\delta t^2}\right)=(\alpha x+\beta y^2)\frac{\partial}{\partial t}\left(\frac{1}{\gamma z+\delta t^2}\right)=(\alpha x+\beta y^2)\frac{-1}{(\gamma z+\delta t^2)^2}\frac{\partial}{\partial t}(\gamma z+\delta t^2)=-\frac{2\delta t(\alpha x+\beta y^2)}{(\gamma z+\delta t^2)^2}.$$
