Answer
$$\frac{{\partial w}}{{\partial t}} = - \frac{{\cos t\sin t}}{{\sqrt {1 + {{\cos }^2}t} }}$$
Work Step by Step
$$\eqalign{
& w = \sqrt {{x^2} + {y^2} + {z^2}} ,\,\,\,\,\,\,\,x = \sin t,\,\,\,\,\,\,\,\,y = \cos t,\,\,\,\,\,\,\,\,z = \cos t \cr
& \cr
& {\text{Calculate the partial derivatives }}\frac{{\partial w}}{{\partial x}},\,\,\,\frac{{\partial w}}{{\partial y}},\,\,\frac{{\partial w}}{{\partial z}} \cr
& \frac{{\partial w}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\sqrt {{x^2} + {y^2} + {z^2}} } \right] = \frac{{2x}}{{2\sqrt {{x^2} + {y^2} + {z^2}} }} = \frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }} \cr
& \frac{{\partial w}}{{\partial y}} = \frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }} \cr
& \frac{{\partial w}}{{\partial z}} = \frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }} \cr
& \cr
& {\text{Calculate the derivatives }}\frac{{dx}}{{dt}},\,\,\,\frac{{dy}}{{dt}},\,\,\frac{{dz}}{{dt}} \cr
& \frac{{dx}}{{dt}} = \cos t,\,\,\,\,\,\,\frac{{dy}}{{dt}} = - \sin t,\,\,\,\,\,\,\,\,\,\frac{{dz}}{{dt}} = - \sin t \cr
& \cr
& {\text{Use the chain rule }}\frac{{\partial w}}{{\partial t}} = \frac{{\partial w}}{{\partial x}}\frac{{dx}}{{dt}} + \,\,\frac{{\partial w}}{{\partial y}}\frac{{dy}}{{dt}} + \,\,\frac{{\partial w}}{{\partial z}}\frac{{dz}}{{dt}} \cr
& \frac{{\partial w}}{{\partial t}} = \frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\frac{{dx}}{{dt}} + \,\,\frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\frac{{dy}}{{dt}} + \,\frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\frac{{dz}}{{dt}} \cr
& \frac{{\partial w}}{{\partial t}} = \frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\left( {x\frac{{dx}}{{dt}} + \,\,y\frac{{dy}}{{dt}} + \,z\frac{{dz}}{{dt}}} \right) \cr
& \frac{{\partial w}}{{\partial t}} = \frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\left( {x\cos t - \,\,y\sin t - z\sin t} \right) \cr
& \cr
& {\text{Write in terms of }}t \cr
& \frac{{\partial w}}{{\partial t}} = \frac{1}{{\sqrt {{{\sin }^2}t + {{\cos }^2}t + {{\cos }^2}t} }}\left( {\sin t\cos t - \,\,\cos t\sin t - \cos t\sin t} \right) \cr
& \frac{{\partial w}}{{\partial t}} = - \frac{{\cos t\sin t}}{{\sqrt {1 + {{\cos }^2}t} }} \cr} $$
