Ahora ya está.
$$\begin{align}&u=xy^4+z^3x^2+Ø(y/x)\\&\\&\frac{\partial u}{\partial s}=\frac{\partial x}{\partial s}y^4+x·4y^3 \frac{\partial y}{\partial s}+3z^2 \frac{\partial z}{\partial s}x^2+z^3·2x \frac{\partial x}{\partial s}+\frac{\partial Ø}{\partial \frac yx}·\frac{\partial \frac yx}{\partial s}=\\&\\&2rse^{-t}(1+rse^{t})^4+4rs^2e^{-t}(1+rse^t)^3(re^t)+\\&3r^4s^2 sen^2t·r^2sent·r^2s^4e^{-2t}+2r^6s^3sen^3t·rs^2e^{-t}·2rse^{-t}+\\&\frac{\partial Ø}{\partial \frac yx}·\frac{\partial}{\partial s}\left(\frac{1+rse^{t}}{rs^2e^{-t}} \right)=\\&\\&2rse^{-t}(1+rse^{t})^4+4r^2s^2(1+rse^t)^3+3r^8s^6e^{-2t}sen^3t+\\&4r^8s^6e^{-2t}sen^3t+\frac{\partial Ø}{\partial \frac ys}·\frac{re^t·rs^2e^{-t}-(1+rse^t)2rse^{-t}}{r^2s^4e^{-2t}}=\\&\\&(1+rse^t)^3(2rse^{-t}+2r^2s^2+4r^2s^2)+7r^8s^6e^{-2t}sen^3t+\\&\\&\frac{\partial Ø}{\partial \frac yx}·\frac{r^2s^2-2rse^{-t}-2r^2s^2}{r^2s^4e^{-2t}}=\\&\\&2rs(1+rse^t)^3(e^{-t}+3rs)+7r^8s^6e^{-2t}sen^3t+\frac{\partial Ø}{\partial \frac yx}·\frac{1+2e^{t}}{rs^3e^{-2t}}\\&\\&\text{Y ya es hora de evaluarlo en }(r_0,s_0,t_0)=(2,1,0)\\&\\&\text{Vemos que} \\&\\&Ø(y/x)=Ø\left(\frac{y(2,1,0)}{x(2,1,0)}\right)=Ø (3/2)\\&\\&\text{por lo cual}\\&\\&\frac{\partial Ø}{\partial \frac{y}{x}}\bigg|_{\frac yx=\frac 32}=Ø'(3/2)=-1\\&\\&\text{y la derivada parcial de u respecto de s en ese punto}\\&\\&\frac{\partial u}{\partial s}\bigg|_{(r,s,t)=(2,1,0)}=\\&\\&2·2·1(1+2·1e^0)^3(e^{-0}+3·2·1)+7·2^8·1^6·e^{-2·0}sen^30-1·\frac{1+2e^{0}}{2·1^3·e^{-2·0}}=\\&\\&4·3^3·7+7·256·1·1·0-\frac 3{2}=756+0-\frac 32=\frac{1509}{2}=754.5\\&\end{align}$$
Y eso es todo, espero no haberme equivocadao, repásalo.
Saludos.
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