Total differentiation in a IS curve context
Total differentiation means you are differentiating all variables in the expression. So if we start with
$$y = C\{[y + B - T(y + B)]; [M + \frac{B}{r}]\} + I(r) + G$$
we get:
$$ d y = C_y^{'} dy + C_y^{'} dB - C_y^{'} dTdy - C_y^{'} dTdB + C_A^{'} dM + C_A^{'} \left( \frac{dBr-Bdr}{r^2} \right) + I'dr + dG $$
Here you are using just basic calculus such as chain rule accordding to which $\frac{d}{dx}[y(u(x))] = \frac{dy}{du} \frac{du}{dy}$, plus you use the rule that with total differentiation  $dy(x_1,x_2) = y_1' dx_1 +y_2'dx_2$ and also quotient rule $\frac{d}{dx}[F(x)/G(x)]= \frac{F'G(x) - F(x)G'}{G(x)^2}$.
Next you just move everything to the LHS of the equation to get
$$dy - C_y^{'} dy - C_y^{'} dB + C_y^{'} T'dy + C_y^{'} T'dB - C_A^{'} dM - C_A^{'} \left( \frac{dBr-Bdr}{r^2} \right) - I'dr - dG = 0$$
Now just collect terms and clean up the mess above and you directly get:
$$(1 + C_y^{'}(1 -  T'))dy - (I_r - C_A \frac{B}{r^2})dr - [C_y (1 - T') + \frac{C_A}{r}]dB- C_A^{'} dM  - dG = 0$$
The expression looks difficult because it has a lot of terms but you should not get intimidated by its look.