In my textbook it says write the unknown function of two variables as a product of two functions of a single variable u(x, t) = X(x) T (t) but then the second step it goes straight away to have T ̈(t)/T(t)= X ′′ (x) / X(x). Does it implies that the outcome of differentiate u(x,t) twice with respect to x is always equals to differentiate u(x,t) twice with respect to t?
Edit:
Thanks for the answer from @ Hans Lundmark. However in the following question why does the solution suggest it is T./((k^2)T)=X"/X rather than T ̈(t)/((k^2)T(t))= X ′′ (x) / X(x)?Does the question gives any hint on that?
As stated in the comments, it depends on the equation that you are solving. Let's work out what happens in the problem in your edit.
Let $u(x,t) = X(x)T(t)$. Then $\frac{\partial u}{\partial t} = X(x) T'(t)$ and $\frac{\partial^2 u}{\partial x^2}= X''(x) T(t)$. Substituting these into the differential equation $$ \frac{\partial u}{\partial t} = k^2 \frac{\partial^2 u}{\partial x^2},$$ we obtain $$X(x)T'(t) = k^2X''(x)T(t).$$ Now, the idea of separation of variables is to manipulate this equation so that only $x$ variables are on one side (say the right side) and only $t$ variables are on the other side (say the left). So, we will divide both sides by $X(x)$ and also divide both sides by $k^2 T(t)$. This gives $$\frac{T'(t)}{k^2 T(t)} = \frac{X''(x)}{X(x)}.$$