Proof:

To find the derivative of y with respect to x, we need to express y in terms of x and then differentiate. Let's start by solving the equation x = at^3 for t:

x = at^3
t^3 = x/a
t = (x/a)^(1/3)

Now substitute this value of t in the equation y = bt^2:

y = b((x/a)^(1/3))^2
y = b(x/a)^(2/3)

So, we have expressed y in terms of x. Now, let's differentiate y with respect to x to find dy/dx:

dy/dx = d/dx [b(x/a)^(2/3)]

To simplify this expression, let's first use the chain rule. Let u = x/a, so that we can rewrite y as:

y = bu^(2/3)

Now, differentiate y with respect to u:

dy/du = d/dx [bu^(2/3)] * du/dx
dy/du = (2/3)bu^(-1/3) * (1/a)
dy/du = (2b/3a)(x/a)^(-1/3)

Now, we need to find du/dx. Recall that u = x/a:

du/dx = d/dx [x/a]
du/dx = 1/a

Substituting this value back into the expression for dy/du:

dy/du = (2b/3a)(x/a)^(-1/3) * (1/a)
dy/du = (2b/3a^2)(x/a)^(-1/3)

Finally, we can express dy/dx in terms of x:

dy/dx = (2b/3a^2)(x/a)^(-1/3) / (1/a)
dy/dx = (2b/3a^2)(x/a)^(-1/3) * a
dy/dx = (2b/3a)(x/a)^(-1/3)

Now, let's find the third derivative of y with respect to x, which is d^3y/dx^3:

(d^3y/dx^3) = d^3/dx^3 [(2b/3a)(x/a)^(-1/3)]

To differentiate this expression, we can use the power rule repeatedly. Let's start by differentiating (x/a)^(-1/3):

d/dx [(x/a)^(-1/3)] = (-1/3)(x/a)^(-4/3) * (1/a)
d/dx [(x/a)^(-1/3)] = (-1/3a)(x/a)^(-4/3)

Now, let's differentiate again:

d^2/dx^2 [(x/a)^(-1/3)] = d/dx [(-1/3a)(x/a)^(-4/3)]
d^2/dx^2 [(x/a)^(-1/3)] = (-4/3a)(-1/3)(x/a)^(-7/3) * (1/a)
d^2/dx^2 [(x/a)^(-1/3)] = (4/9a^2)(x/a)^(-7/3)

Finally, let's differentiate for the