i used this property
$\displaystyle \frac{\partial}{\partial t}[ X(x)T(t) ] = X(x)[\frac{\partial}{\partial t}T(t)] $ because X(x) won't vary with t
i applied it and still didnt get the needed expresion
http://i50.tinypic.com/67uyc7.jpg
Your equation is originally,
$\displaystyle i h\frac{\partial \Psi}{\partial t}= \frac{h^2}{2m}\frac{\partial^2\Psi}{\partial x^2}$
If you let $\displaystyle \Psi= X(x)T(t)$, you have $\displaystyle \frac{\partial\Psi}{\partial t}= X(x)\frac{dT}{dt}$ and $\displaystyle \frac{dX}{dx}T(t)$ (since X and T are functions of a single variable you should not have "$\displaystyle \partial $") so your equation becomes
$\displaystyle i h X(x)\frac{dT}{dt}= \frac{h^2}{2m}T\frac{d^2X}{dx^2}$
which is what you have. Now divide both sides by XT:
$\displaystyle \frac{i h}{T}\frac{dT}{dt}= \frac{h^2}{2m X}\frac{d^2X}{dx^2}$
the equation you want to get to.
The importance of this, by the way, is the fact that the right side now depends only of x while the left side depends only on t. In order to be equal for all x and t, each must be a constant! That allows you to separate the partial differential equation into two ordinary differential equations,
$\displaystyle \frac{ih}{T}\frac{dT}{dt}= \lambda$
and
$\displaystyle \frac{h^2}{2m X}\frac{d^2X}{dx^2}= \lambda$
which are the same as
$\displaystyle ih \frac{dT}{dt}= \lambda T$
and
$\displaystyle \frac{h^2}{2m}\frac{d^2X}{dx^2}= \lambda X$.
Of course, you have to determine possible values for $\displaystyle \lambda$ and you usually do that by looking at boundary conditions on X. Notice that $\displaystyle \lambda$ is an "eigenvalue" for both differential operators.