Find an equation of the plane tangent to the following surface at the given points.
\[
z=e^{x y} ;(5,0,1) \text { and }(0,4,1)
\]
The tangent plane at $(5,0,1)$ is $z=$
The tangent plane at $(0,4,1)$ is $z=$

Step 1 ：The equation of a tangent plane to a surface \(z=f(x,y)\) at a point \((a,b,f(a,b))\) is given by: \[z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\] where \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\) respectively.

Step 2 ：In this case, \(f(x,y)=e^{xy}\), so we need to find the partial derivatives \(f_x\) and \(f_y\).

Step 3 ：The partial derivative of \(f\) with respect to \(x\) is \(f_x = y*exp(x*y)\) and the partial derivative of \(f\) with respect to \(y\) is \(f_y = x*exp(x*y)\).

Step 4 ：Now that we have the partial derivatives, we can substitute the points \((5,0,1)\) and \((0,4,1)\) into the equation of the tangent plane to find the equations of the tangent planes at these points.

Step 5 ：Substituting the point \((5,0,1)\) into the equation of the tangent plane, we get \(z=1\).

Step 6 ：Substituting the point \((0,4,1)\) into the equation of the tangent plane, we get \(z=1\).

Step 7 ：Final Answer: The equation of the tangent plane to the surface \(z=e^{xy}\) at the point \((5,0,1)\) is \(z=\boxed{1}\) and at the point \((0,4,1)\) is \(z=\boxed{1}\).