C64 Basic, 77 bytes

0fOa=0to1:fOf=0to1:b=1-a:x=a<>f:t=1-f:y=(faNb=0)<>(baNt=0):ifx=ytHnE:nE
1?a+1

Description

Definitions

• box 1 contains gold: a = 0

• box 1 contains frog: a = 1

• box 2 contains gold: b = 0

• box 2 contains frog: b = 1

• box with false inscription contains gold: f = 0

• box with false inscription contains frog: f = 1

• box with true inscription contains gold: t = 0

• box with true inscription contains frog: t = 1

Rules

• "One box contains an angry frog, the other box gold":

  b = 1 - a
  
  t = 1 - f

• "Either this box contains an angry frog, or the box with a false inscription contains an angry frog, but not both":
  x = a XOR f
  in C64 BASIC:

  x = a<>f

• "Either this box contains gold and the box with a false inscription contains an angry frog, or this box contains an angry frog and the box with a true inscription contains gold":
  y = ((b=0) AND f) XOR (b AND (t=0))
  in C64 BASIC:

  y = (f and b=0)<>(b and t=0)

• "one, and only one, of the inscriptions is true":
  x XOR y = 1
  in C64 BASIC:

  x<>y = -1

Algorithm

• iterate through all cases until x<>y = -1
• output box number with gold:
  if box 1 contains gold (a = 0) print 1
  if box 1 contains frog (a = 1) print 2
  so the output is always a + 1

Result

The output is 2, so box 2 contains gold.

Additum

The above program stops at the first result found. But if you iterate to the end, it finds two cases, both with gold in box 2 but different statements about the inscription.

• case 1: gold in box 2, inscription 1 true, inscription 2 false
• case 2: gold in box 2, inscription 1 false, inscription 2 true

Try in emulator