Find an equation for the tangent plane to the surface
$$z+5=x{y}^3\cos (z)$$
at the point $(5,1,0)$.

Step 1 ：Given the surface equation $z+5=x{y}^3\cos (z)$, we want to find the equation of the tangent plane at the point $(5,1,0)$.

Step 2 ：The equation of the tangent plane to a surface given by $f(x,y,z)=0$ at a point $(x_0,y_0,z_0)$ is given by $f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0)=0$, where $f_x$, $f_y$, and $f_z$ are the partial derivatives of $f$ with respect to $x$, $y$, and $z$, respectively.

Step 3 ：First, we need to find the partial derivatives of the function $f(x,y,z)=x{y}^3\cos (z)+5-z$.

Step 4 ：The partial derivative of $f$ with respect to $x$ is $y^3\cos (z)$, with respect to $y$ is $3x{y}^2\cos (z)$, and with respect to $z$ is $-x{y}^3\sin (z)-1$.

Step 5 ：Substituting the point $(5,1,0)$ into the partial derivatives, we get $f_x=1$, $f_y=15$, and $f_z=-1$.

Step 6 ：Substituting these values into the equation of the tangent plane, we get the final equation $x+15y-z-20=0$.

Step 7 ：So, the equation of the tangent plane to the surface $z+5=x{y}^3\cos (z)$ at the point $(5,1,0)$ is $\boxed{{\displaystyle x+15y-z-20=0}}$.