That can be interpreted as either (or both) of two questions:
1) How do we know that
\dfrac{\partial^2 y}{\partial x^2}= \frac{1}{c^2}\dfrac{\partial^2y}{\partial t^2}
(the "wave equation) does describe wave motion with wave speed "c"?
2) How do we show that the movement of a string, under tension T and with linear density \delta satisfies the wave equation with c= T/\delta?
To answer the first one, suppose that f(x, t) satisfies the wave equation. Let g(x, t)= f(x- ct_0, t- t_0). Show that g(x, t) also satisfies the wave equation. Of course, g(x, t) is just f(x, t) with time shifted by t_0 and distance shifted by c t_0 so that the difference in distances is ct_0 for a change in time t_0 so that the speed is ct_0/t_0= c.
Before answering the second, I have to admit it isn't exactly true! It is an approximation but a very good approximation for small waves. If you were to pull on the string or wire really hard, you might well permanently stretch or even break it!
Imagine a small part of the string between (x_0, y(x_0)) and x_0+ \Delta x, y(x_0+\Delta x)). Let the angle the string makes at x_0 be \theta_{x_0}. With tension, T, the upward force at x_0 is T sin(\theta_{x_0}. Let the angle at x_0+ \Delta x be \theta_{x_0+ \Delta x}. The upward force at x_0+ \Delta x is T sin(\theta_{x_0+ \Delta x} so the net force is T(sin(\theta_{x_0+ \Delta x})- sin(\theta_{x_0})). For small angles sine is approximately tangent so we can approximate that by T(tan(\theta_{x_0+ \Delta x}- tan(\theta_{x_0})).
The tangent of the angle of a curve, at any point, is the derivative with respect to x so can write that as
T(\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0}
If that section of string has length s and density \delta then it has mass \delta \Delta s. For small angles, we can approximate \Delta s by \Delta x. "Mass times acceleration" is
\delta \Delta x \dfrac{\partial y^2}{\partial t^2}
so that "force equals mass times acceration" becomes
T(\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0})= \delta \Delta x \dfrac{\partial^2 y}{\partial t^2}
Dividing both sides by T \Delta x, that becomes
\dfrac{\left[\partial y/\partial x\right]_{x_0+\Delta x}- \left[\partial y/\partial x\right]_{x_0}}{\Delta x}= \dfrac{\delta}{T}\dfrac{\partial^2 y}{\partial t^2}
Now we can recognize that fraction on the left as a "difference quotient" which, in the limit as \Delta x goes to 0, becomes the derivative with respect to x. Since the function in the difference quotient is the first derivative of y with respect to x, the limit gives the second derivative:
\dfrac{\partial^2y}{\partial x^2}= \dfrac{\delta}{T}\dfrac{\partial^2 y}{\partial t^2}
the wave equation with 1/c^2= \delta/T so that c^2= T/\delta.
