As the questions says, I have one multivariable function $f(x,y)$ where $x=t*sin(s)$ and $y=t*cos(s)$ I want to calculate $\dfrac {\delta^2} {\delta s \delta t} f(x, y)$
The answer is $f_1*cos(s)-f_2*sin(s)+f_{11}*t*cos(s)*sin(s)+f_{12}*t(cos^2(s)-sin^2(s))-f_{22}*t*sin(s)* cos(s) $
From what I understand I need to first compute $\dfrac{\delta}{\delta t}$ and then from this result compute another partial deriviate with $\dfrac{\delta}{\delta s}$. This should then be $\dfrac{\delta}{\delta s}(f_1*sin(s)+f_2*cos(s))$, but I can't get any further than this on my own.There was one question which asked this very question before, but I didn't get any smarter from looking at the answers they gave, would really appreciate of someone could go through a step by step on how to solve this one! And reason as to why they are doing this, so I get a feel for solving my own ones in the future.
You correctly used the chain rule for differentiation of composite functions in your first step. Now you obtained an expression involving the derivative of a product.
In order to expand it, you must then use the product rule on each term of the sum, so that differentiation must now be applied on each factor independently. For the first term, this gives: $$ \frac{d}{ds} (f_1(x(s, t), y(s, t)) \times \sin(s) ) = f_1(x(s, t), y(s, t)) \times \frac{d( \sin(s) )}{ds} + \frac{d (f_1(x(s, t), y(s, t)) )}{ds} \times \sin(s) $$
Each of the derivatives can now be computed more easily: the first is a well-known formula, and the second involves the chain rule, which you correctly applied for your first result: $$\frac{d (f_1(x(s, t), y(s, t)) )}{ds} = \frac{dx}{ds} f_{11} + \frac{dy}{ds} f_{12}.$$
Expanding the second term is done in the same manner, first by using the product rule and then the chain rule.
The key principle here is to try and isolate the differentiation operators, by progressively splitting them into manageable elements. That is the goal when using differentiation rules.