I have the function $u=xy+\ln{y^2}$ where $x=f(st)$ and $y=g(s+t)$ i have to find $u_s(-1,1)$ knowing that $g(0)=g'(0)=2,\ f(-1)=1$ and $f'(-1)=-1$.
So long this is what i have done: $$u_s=\frac{du}{dx}\frac{dx}{ds}+\frac{du}{dy}\frac{dy}{ds}$$ $$u_s=y \frac{dx}{ds}+\Bigr(x+\frac2 y\Bigr)\frac{dy}{ds}$$ $$u_s=g(s+t)f_s(st)t+\Bigr(f(st)+\frac{2}{g(s+t)}\Bigr)g_s(s+t)$$ Is this right? And... how do i find $u_s(-1,1)$ exactly?
Since $f$ and $g$ are function of one variable, you should write
$$u_s=g(s+t)f'(st)t+\Bigr(f(st)+\frac{2}{g(s+t)}\Bigr)g'(s+t)$$
and then plug in the values you know:
$$g(-1+1)=g(0)=g'(-1+1)=g'(0)=2$$
$$f(-1\cdot1)=f(-1)=1$$
$$f'(-1\cdot1)=f'(-1)=-1$$