If you have $y'(t) = -\lambda y(t)$, then you have a linear ordinary differential equation. You have tagged this question "numerical analysis"; as a consequence, I will address this from a numerical standpoint.
Consider an explicit one-step method. One-step methods can be organized by "order"; a first-order one step method uses a single approximation of the solution in the forward direction; a second-order method essentially divides the interval into two segments, and computes the forward step from there. The most common one-step method is the 4th order Runge-Kutta method, which has a fairly straightforward form.
For all of these methods, there is a region of absolute stability. That is, if you have a linear ODE, if the eigenvalues times the step size $h$ fall within this region, then you are guaranteed to converge on a solution.
You have not specified which method you are asked to analyze. However, it is clear to see that the eigenvalue of your problem is simply $\lambda$; therefore, $\lambda$ must fall within the region of absolute stability for your method.
All explicit one-step methods completely enclose the region of absolute stability for lower-order methods. For Euler's explicit method, the first-order technique, the region of absolute stability is a disk in the complex plane with radius 1 centered at $z = -1$.
Therefore, $h\lambda$ must be within this disk.