Answer
$$\frac{\partial G(x,y,z)}{\partial x}=\frac{x}{(\sqrt{1-x^2-y^2-z^2})^3}$$
$$\frac{\partial G(x,y,z)}{\partial y}=\frac{y}{(\sqrt{1-x^2-y^2-z^2})^3}$$
$$\frac{\partial G(x,y,z)}{\partial z}=\frac{z}{(\sqrt{1-x^2-y^2-z^2})^3}$$
Work Step by Step
We will use chain rule to find these partial derivatives.
The partial derivative with respect to $x$ is:
$$\frac{\partial G(x,y,z)}{\partial x}=\frac{\partial }{\partial x}\left(\frac{1}{\sqrt{1-x^2-y^2-z^2}}\right)=-\frac{1}{2(\sqrt{1-x^2-y^2-z^2)}^3}\frac{\partial }{\partial x}(1-x^2-y^2-z^2)=-\frac{1}{2(\sqrt{1-x^2-y^2-z^2})^3}\cdot(-2x)=\frac{2}{(\sqrt{1-x^2-y^2-z^2})^3}$$
$y$ and $z$ are treated as a constant.
The partial derivative with respect to $y$ is:
$$\frac{\partial G(x,y,z)}{\partial y}=\frac{\partial }{\partial y}\left(\frac{1}{\sqrt{1-x^2-y^2-z^2}}\right)=-\frac{1}{2(\sqrt{1-x^2-y^2-z^2})^3}\frac{\partial }{\partial y}(1-x^2-y^2-z^2)=-\frac{1}{2(\sqrt{1-x^2-y^2-z^2})^3}\cdot(-2y)=\frac{y}{(\sqrt{1-x^2-y^2-z^2})^3}$$
$x$ and $z$ are treated as a constant.
The partial derivative with respect to $z$ is:
$$\frac{\partial G(x,y,z)}{\partial z}=\frac{\partial }{\partial z}\left(\frac{1}{\sqrt{1-x^2-y^2-z^2}}\right)=-\frac{1}{2(\sqrt{1-x^2-y^2-z^2})^3}\frac{\partial }{\partial z}(1-x^2-y^2-z^2)=-\frac{1}{2(\sqrt{1-x^2-y^2-z^2})^3}\cdot(-2z)=\frac{z}{(\sqrt{1-x^2-y^2-z^2})^3}$$
$x$ and $y$ are treated as a constant.
