Can normal chain rule be used for total derivative? Chain rule states that $$\frac{df}{dx} \frac{dx}{dt} = \frac{df}{dt}$$. 
Suppose that $f$ is function $f(x,y)$. In this case, would normal chain rule still work?

The multivariable chain rule goes like this: 
$$
\frac{df}{dt} = \frac{\partial f}{\partial x}\cdot \frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt} 
$$
If you can isolate for $\dfrac{dy}{dx}$, then you can always just do implicit differentiation. 
Let's do an example:
$$
f=f(x,y) = x^2 - y
$$
Where 
$$
x(t) = t, \; \; y(t) = t
$$
$$
\frac{df}{dt} \frac{df}{dx} \cdot \frac{dx}{dt} = 2x -\frac{dy}{dx}\frac{dx}{dt}= 2t - 1
$$
Also
$$
 \frac{df}{dt} = \frac{\partial f}{\partial x}\cdot \frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt} = 2x - 1 = 2t - 1
$$
Now let's do the same example, except this time:
$$
f=f(x,y)= x^2 - y\\
x(t)=t\ln(t), \;\; y(t) = t\sin(t) \\ \Rightarrow
\frac{df}{dt} = 2t\ln(t)(\ln(t) + 1) -(\sin(t) +t\cos(t))
$$
Would implicit differentiation work here? Yes. But  it requires more steps.
Implicit differentiation would be the inefficient route as, 
$$
e^{x/t} = t \Rightarrow y= e^{x/t}\sin(t) \Rightarrow \frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} = \\ \left(\frac{1}{t}\frac{dx}{dt}-\frac{x}{t^2}\right)e^{\frac{x}{t}}\sin(t)+e^{\frac{x}{t}}\cos(t) = \\
e^{\frac{x}{t}}\left(\left(\frac{1}{t}\frac{dx}{dt}-\frac{x}{t^2}\right)\sin(t)+\cos(t)\right) =\\
t\left(\frac{\ln(t)+1}{t} - \frac{t\ln(t)}{t^2}\right)\sin(t) + t\cos(t) = \\
\left(\ln(t) +1 - \ln(t)\right)\sin(t) + t\cos(t) = \\
\sin(t)+t \cos(t)  
$$
But it still works out. 
You can check out this site:
Link 
For examples. 
