We have $S=\{\left(x,y,z\right)\in\mathbb R^3:x^2+y^2+z^2=1,z>0\}$
$C=\{\left(x,y\right)\in\mathbb R^2:x^2+y^2=1\}$ and $\varphi(x,y,z)=x^2+y^3+z^4$
We have to evaluate: $\int_c\nabla\varphi.\tau\, ds$ , $\color {blue}{\text{where $\tau$ is the unit tangent vector to $C$ in the $xy$-plane pointing left as we move clockwise along $C$. }}$
MY TRY:I know the procedure to solve the integral but I just stuck for $\color{red}{\tau}$. I don't know how to deal with it.
I'm going to assume $s$ is the arclength of $C$. In this case, first parameterise $C$ by arclength to obtain a function $\gamma: [0,2\pi] \to S$ whose image is $C$ (can you do this?). Then the function $\varphi(\gamma(s))$ is an ordinary happy function from $[0,2\pi]$ to $\mathbb{R}$, yes? It has a derivative, given by the chain rule: $$\frac{d}{ds}\varphi(\gamma(s))=\nabla \varphi \cdot \tau$$ where $\nabla \varphi$ is evaluated at the point $\gamma(s)$, and $\tau$ is evaluated at $s$. Note the similarity between this derivative and the integrand in your question!
Remark about orientation: when you find $\gamma$ you can choose whether to go round the circle clockwise or anticlockwise. This will influence whether your tangent vector is $\tau$ or $-\tau$. Again, I leave this to you to think about (but tell me if you can't do it!)
Remark about the chain rule: For a function $f(x_{1},\ldots,x_{n})$ of $n$ variables, the gradient of $f$ is defined to be the vector of partial derivatives $$\nabla f = \left(\frac{\partial f}{\partial x_{1}}, \ldots ,\frac{\partial f}{\partial x_{n}}\right)$$ where we obtain the partial derivative $\frac{\partial f}{\partial x_{i}}$ by differentiating with respect to $x_{i}$, while keeping the other variables constant. However, we can choose to let the $x_{i}$ vary in a particular way, rather than in dependently. For example, in your curve $C$, we choose $x,y$ to be related by $x^{2}+y^{2}=1$, or equivalently that $x=\cos\theta$ and $y=\sin\theta$ for $\theta \in [0,2\pi]$. Formally, we have composed the function $\varphi(x,y,z)=x^{2}+y^{3}+z^{4}$ with the function $\gamma(s)=(\cos(s),\sin(s),0)$. So what happens if I change $s$? Well, $\varphi$ changes. How does $\varphi$ change? Well, it depends on how $x,y,z$ change. We intuitively expect that if $x,y,z$ change by $\delta x, \delta y, \delta z$ respectively, then $$\varphi(x+\delta x,y+\delta y,z+ \delta z) \approx \frac{\partial \varphi}{\partial x}\delta x + \frac{\partial \varphi}{\partial y}\delta y + \frac{\partial \varphi}{\partial z}\delta z$$ Now, how do $x,y,z$ change? Well, we have taken $x=\cos(s)$, $y=\sin(s)$ and $z=0$. So we expect $$\delta x \approx -\sin(s) \delta s$$ $$\delta y \approx \cos(s) \delta s$$ $$\delta z =0$$ In other words, if we write $\dot{\gamma}(s)=\frac{d \gamma}{ds}$, then we expect $$\frac{d}{ds} \varphi(\gamma(s))=\nabla \varphi \cdot \dot{\gamma}(s)$$ which is the chain rule (what I have written is not a proof, but given the nature of the question I doubt a proof would be very helpful). For a curve $\gamma$ which is parameterised by arc-length, it turns out that $\dot{\gamma}$ is a unit vector tangent to $\gamma$ at $\gamma(s)$ (it is always a tangent vector, but it might not have unit length if you use a different parameterisation!). In other words, it is either $\tau$ or $-\tau$ (you have to check). From here the curve is a normal one-dimensional integral that you can easily do.
Since $C$ is a circle in 2 dimensions you know its normal vector by computing the gradient of the function $x^2+y^2-1=f(x,y)$. In two dimensions it holds the relation
$n = \nabla f(x,y) / |\nabla f(x,y)|$
and moreover
$\tau = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}n$ (ensures that $n.\tau = 0$)