I’ll give the answer as clearly as I can. I think it will be a long answer.
Going from past experience with a similar question, I’ll write z in terms of t and s by substitution of x and y. Limitations of the system make it impractical to use photoscans because uploads are restricted, and editing becomes a nightmare as it gets progressively more sluggish. There’s also a restriction on the amount of text in an answer.
z=28x²+48xy-10y².
z=28(cosh(5t)cos(3s))²
+48(cosh(5t)cos(3s)sinh(5t)sin(3s))
-10(sinh(5t)sin(3s))².
Take the middle term:
48(cosh(5t)cos(3s)sinh(5t)sin(3s)).
Rewrite it:
48(sinh(5t)cosh(5t)sin(3s)cos(3s)).
We now use trig and hyp trig identities to simplify the expression:
sinh(10t)≡2sinh(5t)cosh(5t), so sinh(5t)cosh(5t)≡½sinh(10t).
sin(6s)≡2sin(3s)cos(3s), so sin(3s)cos(3s)≡½sin(6s).
(I use ≡ to signify an identity—always true.)
The middle term becomes 48(½sinh(10t)(½sin(6s)).
This simplifies to 12sinh(10t)sin(6s).
Therefore:
z=28cosh²(5t)cos²(3s)+12sinh(10t)sin(6s)-10sinh²(5t)sin²(3s).
∂z/∂t can now be found using the chain rule. For partial differentiation, we treat s as a constant:
z=[28cos²(3s)]d(cosh²(5t))/dt+[12sin(6s)]d(sinh(10t))/dt-[10sin²(3s)]d(sinh²(5t))/dt.
I’ve put the constants inside square brackets for clarity. Because we’ve separated the variables t and s we can use d/dt instead of ∂/∂t.
To differentiate we can first write u=cosh(5t) and p=u². du/dt=5sinh(5t) and dp/du=2u.
So:
d(cosh²(5t))/dt=dp/dt=(dp/du)(du/dt)
=2u.5sinh(5t)=10usinh(5t)=10sinh(5t)cosh(5t).
But sinh(5t)cosh(5t)≡½sinh(10t), so:
d(cosh²(5t))/dt=10(½sinh(10t))=5sinh(10t).
And d(sinh(10t))/dt=10cosh(10t).
Now, cosh²(5t)-sinh²(5t)≡1 so sinh²(5t)≡1+cosh²(5t), so the derivatives with respect to t are the same, that is, they both equal 5sinh(10t).
d(sinh²(5t))/dt=5sinh(10t).
We can now piece together ∂z/∂t:
∂z/∂t=(28cos²(3s))(5sinh(10t))+(12sin(6s))(10cosh(10t))-(10sin²(3s))(5sinh(10t)).
We need to evaluate this at (t,s)=(ln(2)/10,π/3).
3s=π so sin(3s)=sin(π)=0, sin(6s)=sin(2π)=0, and cos(3s)=cos(π)=-1. Any terms containing sin(3s) or sin(6s) will be zero and can be ignored, so at the given (t,s):
∂z/∂t=28(5sinh(10t))=140sinh(10t).
But 10t=ln(2). Sinh(ln(2))=½(e^ln(2)-e^-ln(2))=½(2-½)=½(3/2)=¾.
∂z/∂t=140(¾)=105.
∂z/∂s=[28cosh²(5t)]d(cos²(3s))/ds+[12sinh(10t)](6cos(6s))-[10sinh²(5t)]d(sin²(3s)).
Let u=sin(3s), du/ds=3cos(3s), p=u², dp/du=2u.
d(sin²(3s))=dp/ds=(dp/du)(du/ds)=2u.3cos(3s)=6ucos(3s)=6sin(3s)cos(3s)=3sin(6s).
cos²(3s)≡1-sin²(3s),
so differentiating both sides: d(cos²(3s))=-d(sin²(3s))=-3sin(6s).
∂z/∂s=28cosh²(5t)(-3sin(6s))+72sinh(10t)cos(6s)-10sinh²(5t)(3sin(6s)).
∂z/∂s=-84cosh²(5t)sin(6s)+72sinh(10t)cos(6s)-30sinh²(5t)sin(6s).
At the given (t,s):
So, because the first and last terms are zero (sin(3s)=sin(6s)=0):
∂z/∂s=72sinh(10t)cos(6s)=72(¾)cos(2π)=72(¾)(1)=54.