(i) I would first suggest writing this function as a product of the functions,
[tex]\displaystyle y = fgh = (4+3x^2)^{1/2} (x^2+1)^{-1/3} \pi^x[/tex]
then apply the product rule. Hopefully it's clear which function each of f, g, and h refer to.
We then have, using the power and chain rules,
[tex]\displaystyle \frac{df}{dx} = \frac12 (4+3x^2)^{-1/2} \cdot 6x = \frac{3x}{(4+3x^2)^{1/2}}[/tex]
[tex]\displaystyle \frac{dg}{dx} = -\frac13 (x^2+1)^{-4/3} \cdot 2x = -\frac{2x}{3(x^2+1)^{4/3}}[/tex]
For the third function, we first rewrite in terms of the logarithmic and the exponential functions,
[tex]h = \pi^x = e^{\ln(\pi^x)} = e^{\ln(\pi)x}[/tex]
Then by the chain rule,
[tex]\displaystyle \frac{dh}{dx} = e^{\ln(\pi)x} \cdot \ln(\pi) = \ln(\pi) \pi^x[/tex]
By the product rule, we have
[tex]\displaystyle \frac{dy}{dx} = \frac{df}{dx}gh + f\frac{dg}{dx}h + fg\frac{dh}{dx}[/tex]
[tex]\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} (x^2+1)^{-1/3} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} (x^2+1)^{-1/3} \ln(\pi) \pi^x[/tex]
[tex]\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} \frac{1}{(x^2+1)^{1/3}} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} \frac{1}{ (x^2+1)^{1/3}} \ln(\pi) \pi^x[/tex]
[tex]\displaystyle \frac{dy}{dx} = \boxed{\frac{\pi^x}{(4+3x^2)^{1/2} (x^2+1)^{1/3}} \left( 3x - \frac{2x(4+3x^2)}{3(x^2+1)} + (4+3x^2)\ln(\pi)\right)}[/tex]
You could simplify this further if you like.
In Mathematica, you can confirm this by running
D[(4+3x^2)^(1/2) (x^2+1)^(-1/3) Pi^x, x]
The immediate result likely won't match up with what we found earlier, so you could try getting a result that more closely resembles it by following up with Simplify or FullSimplify, as in
FullSimplify[%]
(% refers to the last output)
If it still doesn't match, you can try running
Reduce[<our result> == %, {}]
and if our answer is indeed correct, this will return True. (I don't have access to M at the moment, so I can't check for myself.)
(ii) Given
[tex]\displaystyle \frac{xy^3}{1+\sec(y)} = e^{xy}[/tex]
differentiating both sides with respect to x by the quotient and chain rules, taking y = y(x), gives
[tex]\displaystyle \frac{(1+\sec(y))\left(y^3+3xy^2 \frac{dy}{dx}\right) - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = e^{xy} \left(y + x\frac{dy}{dx}\right)[/tex]
[tex]\displaystyle \frac{y^3(1+\sec(y)) + 3xy^2(1+\sec(y)) \frac{dy}{dx} - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = ye^{xy} + xe^{xy}\frac{dy}{dx}[/tex]
[tex]\displaystyle \frac{y^3}{1+\sec(y)} + \frac{3xy^2}{1+\sec(y)} \frac{dy}{dx} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} \frac{dy}{dx} = ye^{xy} + xe^{xy}\frac{dy}{dx}[/tex]
[tex]\displaystyle \left(\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}\right) \frac{dy}{dx}= ye^{xy} - \frac{y^3}{1+\sec(y)}[/tex]
[tex]\displaystyle \frac{dy}{dx}= \frac{ye^{xy} - \frac{y^3}{1+\sec(y)}}{\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}}[/tex]
which could be simplified further if you wish.
In M, off the top of my head I would suggest verifying this solution by
Solve[D[x*y[x]^3/(1 + Sec[y[x]]) == E^(x*y[x]), x], y'[x]]
but I'm not entirely sure that will work. If you're using version 12 or older (you can check by running $Version), you can use a ResourceFunction,
ResourceFunction["ImplicitD"][<our equation>, x]
but I'm not completely confident that I have the right syntax, so you might want to consult the documentation.
